Question

A -11.0 nC point charge and a +22.0nC point charge are 18.0 cm apart on the x-axis.

A)What is the electric potential at the point on the x-axis where the electric field is zero?

B) What is the magnitude of the electric field at the point on the x-axis, between the charges, where the electric potential is zero?

Answer 1

Let us assume q1 is at Origin ,
q1 = -11 nC @ x =0
q2 = +22 nC @ x =18 cm
(A)
We can take ,
q1 = -q
q2 = 2q
Calculating Point where Electric field = 0 ,

Electric Field is given by, k q1/r^2
Let point be X cm from q1 towards left
Distance of charge from q2 = 18.5 cm + x

E = k ( -q/ x^2 + 2q/ (18.5 +x)^2)
q/ x^2 = 2q/ (18.5 +x)^2)
(18.5 +x)^2) = 2x^2
Solving for x
x = +44.663 cm
As x is on left side of q1 therefore, Point where Electric field is 0 is at x = - 44.663 cm

Electric Potential at  x = - 44.66 cm

E.P = - k*q/x + k*2q/(18.5 + x)
E.P = K (-q/44.663 + 2q/ 63.16)
E.P = 8.9 * 10^9 * 10^-9 * 10^2 *(-11/44.663 + 22/ 63.16)
E.P = 91 Volt

(B)
Calculating Point where Electric Potential = 0 ,
Electric Potential is given by, k q/r
Let point be x cm from q1 towards right

Distance of charge from q2 = 18.5 cm - x
E.P = -k*q/x + 2q/(18.5-x)
q/x = 2q/(18.5 -x)
18.5 - x = 2x
x = 18.5/3
x = 6.2 cm

Let point be x cm from q1 towards left
Distance of charge from q2 = 18.5 cm + x
E.P = -k*q/x + 2q/(18.5+x)
q/x = 2q/(18.5 +x)
18.5 + x = 2x
x = 18.5 cm

Electric Potential is zero at x = 6.2 cm & x = -18.5 cm

Eletric Field at these points -
At x = 6.2 cm
E = 8.9* ( -11/ 6.2^2 + 22/ (18.5 -6.2)^2) N/C
E = 3.84 * 10^4 N/C

At x = -18.5 cm
E = 8.9* ( 11/ 18.5^2 - 22/ (18.5 + 18.5)^2) N/C
E = 0.143 * 10^4 N/C