NCERT solution Class 12 electrostatic potential and capacitance

NCERT Solutions

Class 12 Physics

Electrostatic Potential and Capacitance

NCERT Questions

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Q1.

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.

Case 1:

Let q₁ =5 × 10^–8 C

q₂ = +3x10^–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.

∴ V = (1/ 4πε₀) x (q₁/r) + (1/ 4πε₀) x [q₂/ (d -r)].......... (1)

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.

Electrostatic Potential and Capacitance Class 12 Physics NCERT Chapter 2 NCERT Solutions

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