Class 12 Physics Electrostatic Potential And Capacitance
NCERT Solutions Class 12 Physics Potential And Capacitance Chapter 2 is a part of NCERT Solutions Class 12 Physics. Here we have given NCERT Solutions Class 12 Physics Potential And Capacitance Chapter 2 which will help you to understand and for preparation of CBSE examination and other competitive examinations. Prepare for exams with NCERT Solutions Class 12 Physics Potential And Capacitance Chapter 2 and refer to your friends also.
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NCERT Solutions Class 12 Physics Electrostatic Potential And Capacitance
Question : 1
Two charges 5 × 10–8 C and –3 × 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.
Show Answer :
Answer :
Case 1:
Let q1 =5 × 10–8 C
q2 = +3×10–8 C
Distance between the two charges, d =16cm =0.16m
Distance of point P from charge q1 = r
The electric potential (V) at point P will be 0
Therefore, Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.
Where ε0 = Permittivity of free space
For V=0, equation (1) changes to:
0 = (1/ 4πε0) x (q1/r) + (1/ 4πε0) x [q2/ (d -r)]
⇒ (1/ 4πε0) x (q1/r) = – (1/ 4πε0) x [q2/ (d -r)]
⇒ (q1/r) = – (q2)/ (d -r)
⇒ (5 x 10-8)/(r) = – (-3 x 10-8)/ (0.16-r)
⇒ 5 (0.16-r) = 3r
⇒ 0.8 = 8r
⇒ r =0.1m
=10cm
This shows, the potential is 0 at a distance of 10cm from the positive charge between the charges.
Consider a point P on the outside the system of two charges at a distance of s from the negative charge, where potential is 0.
Then potential is given as:
V = ((1)/ (4πε0)) x ((q1)/s)) + ((1)/ (4πε0)) x (q2)/ (s -d))………. (2)
Where ε0 = Permittivity of free space
For V=0, equation (2) changes to:
0 = ((1)/ (4πε0)) x ((q1)/s)) + ((1)/ (4πε0)) x (q2)/ (s -r))
⇒ (1/ 4πε0) x (q1/s) = – (1/ 4πε0) x [q2/ (s -r)]
⇒ (q1/s) = – [q2/ (s -r)]
⇒ (5 x 10-8)/(s) = = – (-3 x 10-8)/ (s – 0.16)
⇒ 5(s – 0.16) = 3s
⇒ 0.8 = 2s
⇒ s= 0.4m
Or s=40cm
This shows, the potential is 0 at a distance of 40cm from the positive charge between the charges outside the system of charges.
Question : 2
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Show Answer :
Answer :
Consider the figure which shows six equal amount of charges at the vertices of the regular hexagon = q
Charge, q = 5 μC = 5 × 10−6 C
Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d = 10 cm
Electric potential at point O,
V = ((1)/4πε0)) x ((6 × q)/ (d))
Where, ε0 = Permittivity of free space and (1/4πε0)
= (9 × 109) Nm2C-2
∴ V = (9 × 109× 6 × 5 × 10−6)/ (0.1)
= 2.7 × 106 V
Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.
Question : 3
Two charges 2 μC and –2 μC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Show Answer :
Answer :
- An equipotential surface is the plane on which total potential is 0 everywhere. This plane is normal to the line AB.
- The plane is located at the mid -point of line AB because the magnitude of charges is the same.
- The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.
Question : 4
A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C 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?
Show Answer :
Answer :
Given:
- Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 1.6 × 10-7 C
Electric field inside a spherical conductor is zero.
This is because if there is field inside the conductor, then charges will move to neutralize it.
- Electric field E just outside the conductor is given by the relation,
E = (q)/ (4πε0 r2)
Where, ε0 = Permittivity of free space
(4πε0 r2) = (9 x 109) Nm2 C-2
Therefore, E = ((1.6 × 10-7 x 9 x 109)/ (0.12)2)
= (105) Nm2 C-2
- Electric field at a point 18 m from the centre of the sphere = E1
Distance of the point from the centre, d = 18 cm = 0.18 m
E1 = (q)/ (4πε0 dr2)
= ((9 x 109) x (1.6 x 10-7)/ (0.18)2)
= (4.4 x 104) N/C
Therefore, the electric field at a point 18 cm from the centre of the sphere is.4 x 104 N/C
Question : 5
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F).
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?
Show Answer :
Answer :
Given:
Capacitor C0 =8pF =8 x 10-12 F
d2 = (d1)/ (2)
Using formula, C = (Kε0A)/ (d)
Therefore, Ck = (Kε0A)/ (d/2)
= [(2 Kε0A)/ (d)]
By dividing,
(Ck / C) = [(2 Kε0A)/ (d)] x (d/ε0A)
=2K
Or CK =2kC0
= (2 x 6 x 8 10-12)
=96 x 10-12
CK =96pF
Mddle block 1
NCERT Solutions Class 12 Physics Electrostatic Potential And Capacitance
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?
Show Answer :
Answer :
- Capacitance of each of 3 capacitors, C =9pF
Equivalent capacitance of the 3 capacitors is:-
(1/C’) = (1/C) + (1/C) + (1/C)
= (3/C)
= (3/9)
= (1/3)
⇒ (1/C’) = (1/3)
⇒ C’ = 3pF
Therefore, total capacitance of the combination is 3 pF.
- Supply voltage, V = 100 V
Potential difference (V1) across each capacitor is equal to one-third of the supply voltage.
∴ V1 = (V/3)
= (120/3)
= 40 V
Therefore, the potential difference across each capacitor is 40 V.
Question : 7
Three capacitors of capacitances 2 pF, 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.
Show Answer :
Answer :
Given:
Capacitances of the given capacitors: C1 = 2 pF, C2 = 3 pF and C3 = 4 pF
For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum,
Therefore Ceq = (C1 + C2 + C3) = (2 + 3 + 4) = 9 pF
Therefore, total capacitance of the combination is 9 pF.
(b) Supply voltage, V = 100 V
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation, q = VC
For C = 2 pF, charge = VC = 100 × 2 = 200 pC = 2 × 10–10 C
For C = 3 pF, charge = VC = 100 × 3 = 300 pC = 3 × 10–10 C
For C = 4 pF, charge = VC = 100 × 4 = 400 pC = 4 × 10–10 C
Question : 8
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 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?
Show Answer :
Answer :
Given:
Area of each plate A= 6 × 10–3 m2
Distance between each plates d= 3mm = 3 x 10-3 m
Using, C = (Aε0)/ (d)
= (6 × 10–3 x 8.85 x 10-12)/ (3 x 10-3) = 1.77 x 10-11 F
Also, q = (CV)
= (1.77 x 10-11 F x 100) =1.77 x 10-9C ≈1.8 x 10-9C
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.
Show Answer :
Answer :
Given:
- Dielectric constant of the mica sheet, k =6
If voltage supply remained connected, voltage between two plates will be constant.
Supply voltage, V = 100V
Initial capacitance, C=1.771 x 10-11F
New capacitance, C1 = kC =6 x 1.771 x 10-11 F= 106pF
New charge, q1 = (C1V) = (106 x100) pC =1.06 x 10-8C
Potential across the plates remains 100V.
- Dielectric constant, k =6
Initial capacitance, C=1.771 x 10-11F
New capacitance, C1 = kC = (6 x 1.771 x 10-11) F= 106pF
If supply voltage is removed, then there will be constant amount of charge in the plates which is (1.771 x 10-11) F
Potential across the plates is given by,
V1 = (q/C1)
= (1.771 x 10-11)/ (106 x10-12)
=16.7V
Question : 10
A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Show Answer :
Answer :
Given:
Capacitor of the capacitance C = 12pF = 12 x 10-12 F
Potential difference, V =50V
Electrostatic energy stored in the capacitor is given as:
E = (1/2) CV2
= (1/2) x 12 x 10-12 x (50)2 J
=1.5 x 10-8 J
The electrostatic energy stored in the capacitor is 1.5 x 10-8 J
NCERT Solutions Class 12 Physics Electrostatic Potential And Capacitance
Question : 11
A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor.
How much electrostatic energy is lost in the process?
Show Answer :
Answer :
Given:
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
E1 = (1/2) CV2
=12× (600 × 10−12) × (200)2J
= 1.2 × 10−5J
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it,
then equivalent capacitance (C’) of the combination is given by,
(1/C’) = (1/C) + (1/C)
= (1/600) + (1/600)
= (2/600)
= (1/300)
Therefore, C’ = 300pF
New electrostatic energy can be calculated as follows:
E2 = (1/2) C’V2
= (1/2) × (300 × (200)2) J
= 0.6 × 10−5J
E2=6 × 10−6J
Therefore, the electrostatic energy lost in the process is 6 × 10−6 J.
NCERT Solutions Class 12 Physics Electrostatic Potential And Capacitance
NCERT Solutions Class 12 Physics Electrostatic Potential And Capacitance
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