Question

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 electricpotential zero? Take the potential at infinity to be zero.

Answer 1

Given: The two charges are 5× 10 −8 C and −3× 10 −8 C, distance between the charges is 16 cm and potential at infinity is zero.

Case (1):

Consider a point P between the line joining two charges where potential is zero.

Potential at point P is given as,

V= q 1 4π ε 0 r + q 2 4π ε 0 ( d−r )

where, potential at point P is V, two charges are q 1 and q 2 , permittivity of free space is ε 0 distance between charges is d and the distance of point P from q 1 is r.

By substituting the given values in the above equation, we get

0= 5× 10 −8 4π ε 0 r + ( −3× 10 −8 ) 4π ε 0 ( 16 cm×( 1 m 100 cm )−r ) 5× 10 −8 4π ε 0 r = 3× 10 −8 4π ε 0 ( 0.16−r ) 0.16 r −1= 3 5 r=0.1 m

Thus, the potential is zero at a distance of 10 cm from q 1 charge.

Case (2):

Consider a point P outside the system of two charges where potential is zero.

Potential at point P is given by,

V= q 1 4π ε 0 r + q 2 4π ε 0 ( r−d )

where, potential at point P is V, two charges are q 1 and q 2 , permittivity of free space is ε 0 distance between charges is d and the distance of point P from q 1 is r.

By substituting the given values in the above equation, we get

0= 5× 10 −8 4π ε 0 r + ( −3× 10 −8 ) 4π ε 0 ( r−16 cm×( 1 m 100 cm ) ) 5× 10 −8 4π ε 0 r = 3× 10 −8 4π ε 0 ( r−0.16 ) 1− 0.16 r = 3 5 r=0.4 m

Thus, the potential is zero at a distance of 40 cm from q 1 charge.