NCERT Solutions for Electrostatic Potential and Capacitance Class 12 Physics
Book Solutions1
Answer
2
A regular hexagon having sides equal to 10 cm, contains charges of 5 µC at its each vertices.
Find the potential at the centre of this regular hexagon.
Answer
3
Charges 2 µC and -2 µC at a point A and B, respectively are 6 cm apart.
(1) Obtain an equipotential surface of the system.
(2) At every point on the surface find the direction of electric field.
Answer
(1) An equipotential surface is defined as the surface over which the total potential is zero. In the given question this plane is normal to line AB. The plane is located at the mid – point of the line AB as the magnitude of the charges are same.
(2) At every point on this surface the direction of the electric field is normal to the plane in the direction of AB.
4
A conductor is in the shape of a sphere with radius 12 cm having charge 1.6 x 10-7C uniformly distributed over its surface. What will be the electric field ?
(1) Inside the sphere.
(2) Just outside the sphere.
(3) At a point 18 cm from the centre of sphere.
Answer
5
Answer
6
Three capacitors connected in series have capacitance of 9pF each.
(1) What can be total capacitance of this combination?
(2) Find the potential difference across each capacitor if they are connected to a 120v supply?
Answer
7
In parallel combination three capacitors of capacitances 2pF, 3pF and 4pF are connected.
(1) What will be the total capacitance of this combination?
(2) If the connection is connected to 100V supply, find the charge on each capacitor.
Answer
1) Given, C1 = 2pF, C2 = 3pF and C3 = 4pF.
Equivalent capacitance for the parallel combination is given by Ceq .
Therefore, Ceq = C1 + C2 + C3 = 2 + 3 + 4 = 9pF
Hence, total capacitance of the combination is 9pF.
(2) supply voltage, V = 100v
The three capacitors are having the same voltage, V = 100v
q = vc
where,
q = charge
c = capacitance of the capacitor
v = potential difference
for capacitance, c = 2pF
q = 100 x 2 = 200pC = 2 x 10-10C
for capacitance, c = 3pF
q = 100 x 3 = 300pC = 3 x 10-10C
for capacitance, c = 4pF
q = 100 x 4 = 400pC = 4 x 10-10 C
8
Answer
9
Considering the same capacitor as in Q 8, when a thick mica sheet, 3mm wide is placed between the plates (Dielectric constant = 6 )
(a)With constant voltage supply
(b) After disconnecting the voltage supply
Answer
10
Answer
Given,
Capacitance of the capacitor, C = 12pF = 12 x 10-12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
E =
Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10-8 J.
was disconnected.
11
Answer
