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 Bihar Board Class 12 Physics Solutions Chapter 2 Electrostatic Potential and Capacitance

Bihar Board Class 12 Physics Electrostatic Potential and Capacitance Textbook Questions and Answers

Question 1.
Two charges 5 x 10-8 C and – 3 x 10-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Answer:
Here, q1 = 5 x 10-8 C, q2 = – 3 x 10-8 C.

r = distance between q1 and q2 = 16 cm = 0.16 m.
Let q1 and q2 lie at points A and B.
Let P be the required point i.e, the point on the line AB at a distance x m from qj at which the electric potential is zero. If V1 and V2 be the potential at P due to q1 and q2 respectively, then using the formula,

If V be the total potential at P, then

on the side of -ve charge from 5 x 10-8 C.

Question .2.
A regular hexagon of side 10 cm has a charge 5 μC at at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer:
Let ABCDEF be a regular hexagon of side 10 cm. O is the centre of the hexagon.
Here, q = charge on each vertex of the hexagon = 5 μC = 5 x 10-6 C.
Sides = AB = BC = CD = DE = EF = FA = 0.10 m

Question 3.
Two charges 2 μC and -2μC are placed at points A and B 6 cm apart.
(a) Identify an equi-potential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Answer:
(a) Here, two charges 2μC and – 2μC are situated at point A and B.
∴ AB = 6 cm = 0.06 m.
For the given system of two charges, the equipotential surface is a plane normal to the line joining points A and B. The plane passes through the mid point C of the line AB. The potential at C is

Thus potential at all points lying on this plane is equal and is zero, so it is an equipotential surface.

(b) We know that the electric field always acts from + ve to – ve charge, thus here the electric field acts from point A (having + ve charge) to point B (having – ve charge) and is normal to the equipotential surface.

Question 4.
A spherical conductor of radius 12 cm has a charge of 1.6 x 10-7 C distributed uniformly on its surface. What is the electric field
(a) inside the sphere,
(b) just outside the sphere,
(c) at a point 18 cm from the centre of the sphere?
Answer:
Here q = charge on the conductor = 1.6 x 10-7C.
r = radius of the spherical conductor = 12 cm = 0.12 m.
(a) We know that the charge given to a spherical conductor resides on its outer surface.
∴ electric field inside the here spherical conductor is zero.

(b) For a point just outside the sphere i.e., for a point lying on the surface of the sphere, the charge may be supposed to be concentrated bn the centre of the sphere.
Thus using the relation,

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (lpiF = 10-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?
Answer:
Here, C0 = capacitance of the parallel plate capacitor with air between the plates = 8pF = 8 x 10-12 F
Let A = Area of its each plate
d = distance between the plates.
Thus,
C0=ε0⋅Ad ………….(i)
Let d’ = distance between the plates with a dielectric substance between them = d2
ε = ε0 K
A = same
C = capacitance of the parallel plate capacitor in presence of the dielectric substance = ?
K = 6
thus,

Question 6.
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Answer:
Here, C1= C2 = C3 = 9 pF = 9 x 10-12 F = capacitance of each of three capacitors.
V = Voltage applied = 120 V.
(a) Cs = capacitance of series combination = ?

(b) Let V1, V2, V3 be the potential difference across each capacitor = ?
∴ Sum of V1, V2, V1 must be equal tc 20 V.
i.e. V1 +V2 + V3 = 120 ………….(1)
Let q = charge on each capacitor.
Also we know that

Now as capacitance of each tapacitor is same, so P-D. across each capacitor must be san^e i.e.,

Question 7.
Three capacitors of capacitances 2 p$, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer:
Here, C1 = 2pF = 2 x 10-12 F, C2 = 3pF = 3 x 10-12 F, C3 = 4pF = 4 x 10-12

V = P.D. applied across the combination.

(a) Cp = Total Capacitance of the parallel grouping = ?
We know that the total capacitance of the parallel combination is given by
C p = C1 + C2 + C3
= 2 x 10-12 + 3 x 10-12 + 4 x 10-12
= 9 x 10-12 F = 9 pF.

(b) Let q2 and q3 be the charges on the capacitors C1, C2 and C3
respectively.
Also we know that in parallel combination, the potential difference across each capacitor
= supply voltage = 100 V.
q1= C1V = 2 x 10-12 x 100 = 2 x 10-10 C.
q2= C2V = 3 x 10-12 x 100 = 3 x 10-10 C.
q3= C2V = 4 x 10-12 x 100 = 4 x 10-10 C.

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 x 103 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Answer:
A = Area of each plate of the parallel plate capacitor = 6 x 10-3 m2.
d = distance between the plates = 3 mm = 3 x 10-3 m.
E0 = 8.854 x 10-12 C2 m2 N-1.
Let C0 = capacitance of the capacitor in presence of air between the plates = ?

Now V0 = P.D. applied across the capacitor = 100 V
q0 = charge on each plate of capacitor = ?
Using the relation, q0 = C0V0, we get
q0 = 17.708 x 10-12 x 100
= 17.708 x 10-10 C = 1.771 x 10-9 C = 1.8 x 10-8 C

Question 9.
Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
Answer:
(a) Here C0 = Capacitance of the capacitor with air as medium = 18 pF .
d – distance between plates = 3 x 10-3 m.
t = thickness oUmica sheet = 3 x 10-3 m = d.
K = dielectric constant of the sheet = 6
As the mica sheet completely fills the space between the plates, thus the capacitance of the capacitor (C) is given by
C = KC0 = 6 x 18 x 10-12 F = 108 x 10-12 F = 108 pF.
Thus the capacitance of the capacitor increases by K times on inserting the mica sheet.
P.D. across this capacitor, V = 100 V
∴ Charge q’ on the capacitor with mica sheet as medium is given by
q’ = CV = 108 x 10-12 x 100 = 108 x 108 U
Now clearly q’ = KC0 V = Kq = 6 x 1.8 x 10-9
= 1.08 x 10-8C.
Clearly charge becomes K times the charge on the plates with air as medium i.e., charge on the plates increases when supply remains connected and mica sheet is inserted.

(b) Here, capacitance of capacitor with mica as medium
C = KC0 = 108 x 10-12 F
When supply is disconnected i.e. m V = 0,
The potential difference across on the plates of the capacitor V’ reduces by K times.
i.e„ , V = 1006 = 1.6.67 V
C becomes 6 times.
Thus if q1 be the charge on its plates after disconnecting the supply,
Then

i.e., the charge on the capacitor with mica as medium remains same as with air medium.

Question 10.
A .12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor? .
Answer:
Here, C = capacitance of the capacitor = 12 pF = 12 x 10-12 F.
V = supply voltage connected across the capacitor = 50 V.
U = electiostatic energy stored in the capacitor Using the relation,
U =12 CV2, we get
U = j x 12 x 10-12 x (50)2
= 6 x 10-12 x 2500 = 150 x 10-12 J = 1.5 x 10-8J.

Question11.
A 600 pF capacitor is charged by a 200 Vsupply.lt is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Answer:
Here, C, = capacitance of capacitor = 600 pF = 600 x 10-12 F = 6 x 1010 F.
V1 = supply voltage = 200 V
Let U1 = initial electrostatic energy stored in the capacitor, then using the relation
U = 12 CV2 we get
U1 = 12 x C1V12 = 12 x 6 x 10-10 x (200)2= 12 x 106 J.
C2= capacity of another capacitor = 600 pF = 6 x 102 F
q2 = charge on second capacitor = 0
q1= charge on first capacitor.
q1 = C1V1 = 6 x 10-10 x 200 = 12 x 108 C.

When first capacitor is connected to another uncharged capacitor, then the charge will be equally shared by the two capacitors.
If q = charge attained by each capacitor on joining, then

Let V be the common potential, then

∴ loss in electrostatic energy = U1 – U2
= 12 x 10-6 x 10
= 6 x 10-6J.

Question 12.
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 x 10-9 C from a point P (0,0, 3 cm) to a point Q (0,4 cm> 0), via a point R (0,6 cm, 9 cm).
Answer:
Here q = charge at origin O.
= 8 mC = 8 x 10-3C.
q0= charge to be carried from P to Q via R
= -2.x 10-9 C.
Initial position vectorof P, r1→ = 3 k^cm.
Final position vector at Q. r2→ = 4j^cm.
∴ r1= 3 cm = 3 x 10-2m
r2 = 4 cm = 4 x 10-2 m
As electrostatic force are conservative forces, the work done in moving q0is independent of the path followed. Thus there is no relevance of the point R.

Question 13.
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
Answer:
Here, side of the cube = b, q = charge on each vertex of the cube.
Now HB = diagonal of the face ABGH
= AB2+AH2−−−−−−−−−√
= b2+b2−−−−−−√

∴ Diagonal HC of the cube is given by,
HC = HB2+BC2−−−−−−−−−−√ = b2+b2+b2−−−−−−−−−√=b3–√
HO = OC = b3√2

Similarly AO = OB = OE = OG = OF = OD = 3√2 b i.e, the distance of each vertex from the centre O of the cube is same = 3√2 b
or distance of each charge from O = 3√2 b.
Let V = Electrical potential at O = ? and E = electric field at O = ?
We know that

‘The electric field at O due to charges at the opposite comers such as A and F, B and E, C and H, D and G are equal in magnitude and opposite in direction. So the net electric field at O is zero due to the symmetry of charges.

Question 14.
Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid-point.
Answer:
Here, q1 = 1.5 μC = 1.5 x 10-6 C.
q2= 2.5 μC = 2.5 x 10-6 C.

r = distance between the two charges = 30 cm = 0.30 m

Let O be the middle point of the line AB.
(a) (i) If V1 and V2be the electric potentials at O due to 1.5 μC and 2.5 μC respectively, then using the relation,

If V be the total potential at O due to the two charges, then

(ii) Electric field at O – Let E⃗  = total electric field at O = ?
If E⃗ 1 and E⃗ 2 be the fields at O due to q1 and q2 respectively, then
using the relation, E =14πε0⋅qr2, we get

along OA i.e., towards q1
or E = 4.0 x 105 Vm-1 From 2.5 μC to 1.5 μC.

(b) Let P be the point at a distance of 10 cm from O in a plane normal to AB.
AO = OB = 0.15 m
OP = 10 cm – 0.10 m.

Let V’ be the total potential at P = ?
Now in rt. ∠d ∆ AOP,

(ii) If E1and E2 be the resultant electric field at P due to q1 and q2, then using the relation, then


If E′→ be the resultant electric field at P due to q1 and q2, then using the relation,
R = P2+Q2+2PQcosθ
we get the magnitude of E⃗  given by

To find direction of E′→ w.r.t. the line AB – Let ∠PAB = β. Then in rt. ∠d ∆ AOP, B + θ2 90°, β = 90° – θ2 = 90 – 56.25° = 33.75° = 33.8 .
∴ angle made by E′→ with AB line = β + α = 33.8′ + 76.9° = 110.7
or angle made by E′→ with the line joining 2.5 µC to 1.5 µC charge i.e., line BA = 180° -110.7 = 69.7°.

Question 15.
A spherical conducting shell of inner radius rx and outer radius r2 has a charge Q.
(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Answer:
(a) We know that any charge given to a hollow conductor spreads over the surface of the conductor and electric field inside the hollow conductor remains zero. Thus the charge Q given to the conducting shell will spread on its outer surface. The another charge q is placed at the centre of the same shell having inner radius rr It will induce – q charge on the inner surface and + qon outer surface which will shift to the outer surface of the shell of radius r2. Thus total charge on outer surface is Q + q. Let A1 and A2 be the area of inner, and outer shell respectively.

(b) Yes. According to Gauss’s law, 
For a cavity of arbitrary shape, this is not enough to claim that E⃗  = 0 inside the cavity. The cavity may have +ve and -ve charges with total charge zero. Charge tends to reside on the outer surface of a conductor. The result of this law is independent of the shape and size of cavity.

Question 16.
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

where n^ is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of n^ is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ n^ /ε0

(b) Show that the tangential components of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a) use Gauss’s law. For (b) use the fact that work done by-electro static field on a closed loop is zero.]
Answer:
(a) Let AB be a charged surface having two sides as marked in the figure. A cylinder enclosing a small area element As of the charged surface is the Gaussian surface.

Let σ = surface charge density
∴ q = charge enclosed by the Gaussian cylinder = σ. ∆s
∴ According to Gauss’s Theorem,

Where E1−→ + E2−→ are the electric fields through circular cross – sections of cylinder at II and III respectively.

Hence, proved.

It is clear from the figure that E1−→ lies inside the conductor. Also we know that the electric field inside the conductor is zero.
∴ E1−→ = 0.
Thus from equation (1),

or electric field just outside the conductor = σε0n^
Hence, proved.

(b) Let AaBbA be a charged surface in the field of a point charge q lying at origin.
Let rA→ and rB→ be its position vectors at points A and B respectively.
Let E→ be the electric field at point P, thus E cos θ is the tangential component of electric field E⃗  .
E→. dl→ = E dl. cos θ = (E cos θ) dl.

To prove that E cos θ is continuous from one to another side of the charged surface, let us find the value of  if comes to be Zero
then we can say that tangential component of E→ is continuous.

Hence, proved.

Question 17.
A long charged cylinder of linear charge density X is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Answer:
Consider two cylindrical Co-axial shells each of length / with outer and inner radii r2 and r1 (r2 >r1). The inner cylinder is charged having linear charge density X and enclosed by a hollow cylinder of radius r2. We want to find out the electric field E⃗  in the space between the two cylinders.

If + q be the charge on the inner cylinder, then q = A.1. Due to electrostatic induction, – q charged is induced on the inner surface of the hollow cylinder and + q charge is induced on its outer surface and is earthed. Let r be the distance from XX’ axis of cylinders of a point P between the cylinders at which E→ is to be. calculated. Draw a cylindrical Gaussian surface of length l, radius rst. The point P lies on its surface. The charge enclosed inside the Gaussian surface is given by
q = λ l.
The electric field E→ at any point on the Gaussian surface will be uniform, equal in magnitude and towards the outward drawn normal due to the spherical symmetry.
Curved surface area of Gaussian surface = 2πrl
Consider an area element dS on the curved surface at point P.
If dø be fhe electric flux through ds→, then by def. dø = E→ .ds→ = E dS cos 0 = E dS
(∴E→ and ds→ acts along n^ )
If ø1 be .the electric flux through ds→the whole curved surface area of the Gaussian surface, then

Now. according to Gauss’s Theorem,
ø = ø1 + ø2 = E. 2πrl + 0 = E. 2πrl
This field is radial and perpendicular to the axis.
This is the required expression.

Question 18.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 A:
(a) Estimate the potential energy of the system in e V, taking the zero of the potential energy at infinite separation of the electron from proton;
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a) ?
(c) What are the answers to (a) and (b) above if the zero of potential l energy is taken at 1.06 A separation?
Answer:
Here, charge on proton, q1 = 1.6 x 10-19 C.
Charge on electron, q2 = – 1.6 x 10-19 C
r = radius of hydrogen atom = distance between q1 and q2 = 0.53 A
= 0.53 x 10-10m.

(a) If zero of potential energy be taken at infinite seperation, then Potential energy = P.E. at ∞ – P.E. atr

(b) Let W – minimum work required to free the electron = ?
Here, K.E in the orbit = 12 [Magnitude of P.E of case (a)]
= 12 x ( + 27.17 eV)
= 13.585eV (∴ K.LoIthesystem is aÌwas + ve)
∴ Total Energy = K.E + P.E.
= 13.58+(- 27.17) = -13.59 eV
= – 13.6eV.
∴ Work required to free the electron = 0 – (- 13.6)
=13.6e V

(c) Potential energy of system at a separation of 1.06 A between the electron and proton

Potential energy of the system when zero of potential energy is taken at 1.06 Å = – 27.17 – (- 13.585) = – 13.585 eV = – 13. 6 eV.
By shifting the zero of the P.E., work required to free the electron is not affected. It continues to be same, being equal to-(-13.6) = + 13.6 eV.

Question 19.
If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion (H+2 ). In the ground state of a (H–2 ), the two protons are separated by roughly 1.5 Å, and the electron is roughly 1A from each proton. Determine the potential energy of the system. Specify your choice of zero of potential energy.
Answer:
Here, q1 = charge on electron = – 1.6 x 10-19C.
q2, q3 = charges on each protons = + 1.6 x 10-19 C.
Separation between two protons i.e. q2 and q3, = r23 = 1.5 Å = 1.5 x 10-10 m.
Separation between proton and electron = r13 = r122 = 1 Å = 10-10 m.

If the zero of potential energy is taken at infinity,

The zero of potential energy is taken to be at infinity.

Question 20.
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Answer:
Here, a, b = radii of two conducting spheres st. a > b.
Let q1= charge on sphere of radius a.
and q2 = charge on sphere of radius b.
If V1 and V2 be the electric potentials on spheres of radii a and b respectively, then

When the two conducting spheres are connected to each other by a wire, then their electric potentials are same, i.e.,
V1 = V2

Now let E1 and E2 be the electric fields at the surfaces of sphere of radii a and b respectively.
To find E1E2 – E1 and E2 are given by


Thus the ratio of electric field of the first to the second conducting sphere is ba
A flat portion may be taken as a spherical surface of large radius and a pointed portion may be t3ken as a spherical surface of small radius.
As ε ∝  thus pointed portion has larger fields than the flat radius one. Also we know that
E = σε0
i.e. E ∝ σ,
thus clearly the surface charge density on the sharp and pointed ends will be large.

Question 21.
Two charges – q and + q are located at points (0, 0, – a) and (0,0, a) respectively.
(a) What is the electrostatic potential at the points (0,0, z) and (x, y,o)?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >>1.
(c) How much work is done in moving a small test charge from the point (5,0,0) to (- 7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Answer:
(a) Here – q and + q are situated at point A (0,0, – a) and B (0, 0,0) respectively.
dipole length = 2a
If p be the dipole moment of the dipole, then p = 2aq
Let P1 (0,0, z) be the point at which V is to be calculated. It lies on the axial line of the dipole.

Now point P2(x, y, 0) lies in XY plane which is normal to the axis of the dipole i.e., lies on the line parallel to the equitorial line on which potential due to the dipole is zero as given below:

Let OP2 = 7,
∴ r = x2+y2−−−−−−√
∴ If V’ be the eìectric potential at P2, then

(b) Let r= distance of the point P from the centre (O) of the dierect
which V is to be calculated. Let ∠POB = 0 i.e. OP makes an angle 0 with
E→ .Also let rj and r2 be the distances of the point P from – q and + q respectively. To find r1, and r2, draw AC and BD ⊥ arc to OP.

In ∆ ACO, OC = a cos θ
and in ∆BDO, OD = a cos θ
Thus if V1 and V2 be the potentials at P due to – q and + q respectively, then total potential V at P is given by



when ra > > or r2 > > a2, then r2 > > a2 cos2θ, so a2 cos2θ that may be negelected
Thus V = 14πε0⋅pcosθr2 ……(2)
Thus we see that the dependence of V on r is of 1r2 type i.e., V ∝1r2

(c) Let W1 and W2 be the work done in moving a test charge q0 from E (5, 0, 0) to F (- 7, 0, 0) in the fields of + q (0, 0, a) and – q (0, 0, – a) respectively.

BE−→ = 5 i^ – ak^ ,
Bf−→ = – 7i^ – ak^ .
Similarly AE−→ = 5 i^ + ak^, and AF = – 7i^ + ak^
or.  BE =25+a2−−−−−−√ BF = 49+92−−−−−−√
AE = 25+a2−−−−−−√, AF = 49+92−−−−−−√
Using the relation,

Answer:
Here, the electric quadrupole’s is made of four charges + q, + q at A and C, – q and – q and B. Point B may be supposed to be the centre of the quadrupole.
Let r be the distance of the pooint P on the. ax is of the quadrupole from point B at which the electric potential is to be calculated.
Clearly AP = AB + BP = a + r
BP = r and CP = BP – BC = r – a.
Let V be the electric pootential at point P due to the quadrupole.
∴ V is given by

Now for large r, r2 > > 1, so r2a2 > > 1, hence 1 maybe neglected.
∴ eqn. (1) reduces to :

Also we know that the electric potential at a point on axial line to an electric dipole is are given by

The electric potential at a pooint due to a single charge q i.e., monopole at a distance r from it is given by .

Thus we conclude that for larger r, the electric potential due to a quadrupole is inversely proportional to the cube of the distance r while due to an electric dipole, it is inversely proportional to the square of r and inversely proportional to the distance r for a monopole.

Question 23.
An electrical technician requires a capacitance of 2 pF in a circuit across a potential difference of 1 kV. A large number of 1 pF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:
Let N be the total number of capacitors used by the technician and arrange them in m rows each row having n capacitors.
N = mn …(i)

C1 = Capacitance of the each capacitor = 1μF.
C = required capacitance of the combination = 2 μF.
Maximum potential difference across each capacitor = 400 V.
Potential difference across the circuit = 1000 V.
= P.D. across each row.
We know that when the capacitors are connected in series, then pot. differences across them gets added.
∴ n capacitors in a row will stand a voltage = 400 x n
400 x n = 1000
n = 1000400 = 2.5.
As n has to be a whole number not less than 2.5), .’. n = 3.
Let C’ = Total capacitance of the capacitors in a row.



Thus he must connect eighteen 1 pF capacitors in 6 parallel rows, each row containing 3 capacitors.

Question 24.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of pF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Answer:
Here, C = Capacitance of a parallel plate
Capacitor = 2F
d = separation between its plates
= 0.5 cm = 5 x 10-3 m .
A = area of its plates = ?
ε0 = 8.854 x 10-12 C2 N-1 m-2
Thus using the relation,

This area is impossible to realise as it is very large. That is why ordinary capacitors are of range much lesser than of pF.

Question 25.
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Here, C1 – C4 = 1000
pF = 100 x 10-12 F.
C2 = C3= 200 pF = 200 x 10-12 F.
C2 and C3 are in series’combina tion.
If C23, be the equivalent capacitance of C2and C3, in series combination, then



Now C4 and C’ are in series. If C be the equivalent capacitance of series combining of C4 and C’, then

Charge and Voltage across each capacitor – Let q1 q2, q3, and q4 be the charges across the capacitors C1, C2, C3and C4 respectively. Also let V1, V2,V3 and V4be the respective voltage across each of them.
Now as C’ is in series with C4, thus charge on C’ is equal to q4.


Now as C2 and C3 have equal capacitances and are connected in series, so charge on each capacitor must be same.

Question 26.
The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by ,. connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field £ between the plates.
Answer:
Here, A = area of each plate of the parallel plate capacitor = 90 cm2 = 90 x 104 m2.
d = distance between plates = 2.5 mm = 2.5 x 10-3 m
V = Supply voltage = 400 V.

(a) U = electrostatic energy stored by the capacitor = ?
Let C = Capacitance of a parallel plate capacitor,

U is given by
U = 12 CV2 = 12 x 3.186 x 1011 x (400)2
= 2.55 x 10-6 J = 2.55 μj.

(b) We know that electrostatic energy (U) stored by the capacitor is given by
U = 12 CV2 ……..(1)
This can be viewed as the energy stored in the electrostatic field between the plates.
Volume of the parallel plate capacitors is given by
V = Area of plates x distance between plates
= A x d ……(ii)
Energy stored per unit volume of the capacitor is called energy denoted by u.

Also we know that ε0Ad
and V = Potential difference = Ed ……(iv)
= magnitude of electric field between the plates x distance
From (iii) and (iv), we get .

Question 27.
A 4 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 pF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Answer:
Here, Cj = 4pF = 4 x 10-6 F.
V1= ‘ supply connected to capacitor = 200V.
If U1 be the initial energy stored in C1 then

When 4 μF capacitor is connected to uncharged capacitor of 2 pF capacity, the capacitors are in parallel and share charge till both acquire a common potential i.e., P.D. across each capacitor become equal.
∴Total charge on two capacitors is .
q = C1V1 + C2V2 = 4 x 10-6 x 200 + 0
= 8 x 10-4 C (q2 = 0)
Let C’ = total capacitance of two capacitors
∴ C’ = C1 + C1 = 4 + 2 = 6μF = 6 x 10-6F.
If V be the common potential, then

Let U be the energy dissipated by C, in the form of heat and e.m. radiation = ?
∴ U = U1 – U2 = 8 x 10-2 – 5.33 x 10-2
= 2.67 x 10-2 J.

Question 28.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2.
Answer:
Let P and P’ be the parallel plates of the capacitor.

A = Area of its each plate. ‘ d = distance between the plates.
Q = magnitude of charge on each plate.
E = magnitude of electric field between the plates acting from P to V
σ = surface charge density of plates,
∴ E = σε0 …………(i)
Let the separation between the plates be increased by a distance dx against the force of attraction F between the plates.
If dω be the work done in doing so, then
dω = F. dx …….(ii)
This work done goes to increase the P.E. of the capacitor.
Now we know that the potential energy per unit volume of the parallel plate capacitor is given by
u = 12ϵ0⋅E2 ……….(iii)
If du be the increase in potential energy of the parallel plate capacitor on increasing distance d by dx, then
du = uA dx = 12 ε0. E2A dx


Hence proved.
The physical origin the factor 12 in this expression lies in the fact that just outside the conductor, the electric field is E and inside it, it is zero. So the average value of the electric field i.e. E2 contributes to the force against which the plates are moved.

Question 29.
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig). Show that the capacitance of a spherical capacitor is given by –
C = 4π80r1r2r1−r2
where r1 and r2 are the radii of outer and inner spheres respectively.

Answer:
A spherical capacitor consists of two concentric spherical shells A and B of radii r2 and r, respectively s.t. r1 > r2 having centro. When charge – Q is given to the inner spherical shell A, it induces charge + Q on the inner surface of the shell B and – Q on its outer surface. As shell B is earthed, – Q charge on its outer surface flows to earth. Draw a Gaussian surface S of radius r and centre O between shells A and B. The electric field inside the shell A and outside the shell B is zero.

i.e., E = 0 for r < r2 and E = 0 for r > r1
Let E be the electric field at a point P on the Gaussian surface.
E = 14πε0⋅αr2
and acts radially in inward direction.
Also charge enclosed inside the Gaussian surface =- Q.
Let dS−→ be the area element at point P on the Gaussian surface.
If dø be the electric flux through dS−→ then,
dø = E→. dS−→ = E. dS cos 180°
= – E. dS …….(ii)
(∴ angle between E and dS =180°) If ø be the total electric flux through the whole Gaussian surface,
then

Where ∮SdS=4πr2 surface area of the Gaussian surface.
‘Now according to Gauss’s Theorem,

Now let V be the potential difference between the two spherical shells A and B. If C be the capacitance of the spherical capacitor, then
C = QV ………..(v)

Let us now find V – In case of parallel plate capacitor, the E→ between the two plates is uniform and the P.D. between the two plates is simply Ed. But in case of spherical capacitor, the electric field between the two spherical shells is not uniform and it varies with distance as

Question 30.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 pC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Answer:
Here, r2 = radius of inner sphere = 12 cm = 12 x 102 m
r1 = radius of outer sphere = 13 cm = 13 x 10-2m
Q = charge on inner sphere = 2.5 pC = 2.5 x 10-6 C
K = dielectric constant of the liquid filled between two spheres = 32.
(a) Let capacitance of capacitor = C = ?
Using the relation for C due to a spherical capacitor,

(b) Let V = potential of the inner sphere. As the outer sphere is earthed, so the potential of the inner sphere is equal to potential difference between the two spheres of the capacitor.

(c) radius of isolated sphere, r = 12 cm = 12 x 10-2 m.
Let C’ be the capacitance of the isolated sphere.
∴ Using the relation,
C’ = 4πε0r we get

Question 31.
Answer carefully:
(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly gwen by Q1⋅Q24πε0r2 where r as the distance
between their centres?
(b) If Coulomb’s law involved 1/r3 dependence (instead of 1 /r2), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Answer:
(a) The expression for the electrostatic force between two charges q1 and q2 i.e.,
F = 14πε0⋅q1q2r2 holds, when the charges q1 and q2 are point charges only. The expression will not be true in case of large conducLing spheres. It is because, when large conducting spheresare brought together, the charge distribution on the two_spheres will not remain uniform.

(b) No, Gauss’s lawould not be true if Coulomb’s law involved 1r3dependence.

(c) The small test charge will travel along the field line passing through that point and accelerates on a straight line path parallel-to eléctric field E→ if E→ is uniform.
It is not necessary that the tešrcharge will move along the field line passing through that point as the field line gives the direction of acceleration and not that of the velocity in general.

(d) We know that the work done by the electric field in moving a unit test charge between two points in the electric field is equal to1h4ne integral of electric field between two points

Also we know that the line iitegral of electric field along a closed path is zero.
i.e., ∮E→⋅dl→ = 0
Thus the work done by the field of nucleus in a complete ciruclar orbit of electron will be zero as the electrostatic force is conservative for which ∮E→⋅dl→ = 0, hence W = 0.
The same is also true for the elliptical orbit Li’., W = O br elliptical orbit.
(d) Aliter – Zero, the work done does not depend upon the nature of the orbit and ∮E→⋅dl→ = 0 for a closed path:
(e) No, electric potential is not discontinuous across the surface of a charged conductor. It is continuous there. E→ may become zero but V remains constant.
(f) A single conductor also possesses capacitance. It is a capacitor having one of its plate at infinity.
(g) Water molecules are polar and have permanent dipole moment, while the molecules of mica being non-polar don’t have permanent dipole moment, hence water has a much greater value of dielectric constant than mica.

Question 32.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 pC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
Answer:
Elere, l = length of the co-axial cylinders = 15 cm = 15 x 10-2 m.
a = radius of inner cylinder = 1.4 cm = 14 x 10-3 m .
b = radius of outer cylinder = 1.5 cm = 15 x 10-3 m.
q = charge on inner cylinder = 3.5 pC = 3.5 x 10-6 C.
C = Capacitance of the system = ?
V = electric potential of the inner cylinder = ?


Since the outer cylinder is earthed, the potential of the inner cylinder will be equal to the potential difference between them.
Thus V is given by the relation,

Question 33.
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What f minimum area of the plates is required to have a capacitance of 50 pF?
Answer:
Here, K = dielectric constant of the material = 3
Voltage rating V = 1 kV = 103 V
C = Capacitance of the parallel plate capacitor
= 50 pF = 50 x 10-12 F.

Dielectric strength = 107Vm-1 = Emax.
Since the electric field strength is not to exceed 10% of the dielectric j strength, thus if E be the electric field, then

Let q = Charge which the capacitor can hold.
ε0 = 8.854 x 10-12C2 N-1 m-2.
A = required area of the plates of capacitor = ?
Let d be the spacing between the plates of the capacitor.
∴ Using the relation, E = Vdwe get
d = VE = 1000106 = 10-3m
Now q = C x Voltage rating
= 50 x 10-12 x 103 = 5 x 108 C
Also we know that the capacitance of the parallel plate capacitor is given by

Question 34.
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
(a) For a constant electric field acting along Z-direction the equipotential surfaces are planes parallel to the XY-plane as shown in figure here.

(b) As the field increases uniformly in magnitude in a given direction the equipotential surfaces for a corresponding fixed potential difference get closer and are parallel to the XY plane.
(c) The equipotential surfaces for a singlg positive charge at origins will be concentric spheres having centre at origin.
(d) The equipotential surfaces for a uniform grid consisting of long equally spaced parallel charged wires in a plane will be of periodically varying shape near grid which slowly reaches the shape of planes parallel to the grid at far off distances.

Question 35.
In a Van de Graff type generator a spherical metal shell is to be 15 x 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 x 107 Vm-1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Answer:
Here, V = potential due to a charged spherical shell
= 15 x 106V.
E = electric field due to a charged shell = 5 x 107 Vm-1
Let r be the required radius of the spherical shell = ?
We know that the electric potential and the electric field charged shell are given by

Question 36.
A, Small sphere of radius r1 and charge q2 is enclosed by a spherical shell of radius r2 and charge q1 Show that if q2 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.
Answer:
Here r1 r2 are the radii of small sphere and the spherical shell respectively. The shell Surrounds the sphere + q1 is the charge on the sphere. + q2 is the charge on the shell. We know that the electric field
inside a conductor is zero i.e., E→ = 0. Thus according to Gauss’s Theorem,

q2 = 0 inside the spherical shell as E→ = 0 inside it.

Hence q2 must reside on the outer surface of the spherical shell. Now the sphere having +q1 charge is enclosed inside the spherical shell. So -q1 charge will be induced on the inside side and + q1 charge will be induced on the outer surface the spherical shell.

∴ Total charge on the outer surface of the shell = q2 + q1 As the charge always resides on the outer surface, thus charge q, from the outer surface of sphere will flow to the other surface of spherical shell when connected with a wire.

Question 37.
Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm-1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be steel cage so there is no field inside!)

(b) A man fixes outside his house one evening a two metre high insulting slab carrying on its top a large aluminium sheet of area 1 m2. Will he get an electric shock if he touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conducting of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
[Hint: The earth has an electric field of about 100 Vm-1 at its surface in the downward direction, corresponding to a surface charge density = -10-9 C m-2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about +1800 C is pumped every second into the earch as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.]
Answer:
(a) Our body and the earth become an equipoJtential surface which means that there is no potential difference between the earth and our body. Hence as we come out from our house into open, original equipotential surfaces of open air change in such a way that the head and ground remain at the same potential. Thus no current flows through our body and therefore we don’t experience an electric shock.

(b) Yes, he will get an electric shock if he touches the metal sheet next morning. This is because the aluminium sheet and the earth form a capacitor with the insulating slab as a dielectric. The down pour of the atmospheric charge will raise the potential of the sheet of aluminium, i.e., it gets charged by the discharging current of 1800 A coming down from the stratosphere. When we touch the aluminium sheet, charge will flow to the earth through our body. This flow of charge constitute an electric current and we will experience a shock.

(c) The atmosphere is continuously being charged by lightning and thunder storms all over the globe and maintains an equilibrium with the discharge of atmosphere during normal weather conditions. Hence the atmosphere cannot become electrically neutral.

(d) The electrical energy of the atmosphere is dissipated during lightning in the form of light, heat and sound during thunder storms.