CBSE Class 12 Physics 2019 Delhi Set 2 Paper

Question : 5 of 6

Marks: +1, -0

(a) Draw equipotential surfaces corresponding to the electric field that uniformly increases in magnitude along with the Z-direction.

(b) Two charges

-q

and

+q

are located at point

(0,0,-a)

and

(0,0, a)

. What is the electrostatic potential at the points

(0,0, \pm z)

and

(x, y, 0)

Solution:

(a)

V_z = \frac{k q}{(z-a)} + \frac{k(-q)}{(z+a)} \text{ if } z>a

=K q \left[\frac{1}{z-a}-\frac{1}{z+a}\right] = K q \frac{2a}{z^2-a^2}

Potential at pt.

(0,0-z)

V_{(-z)} = \frac{K q}{(z-a)} + \frac{K(-q)}{(z+a)}

=K q \left[\frac{1}{z-a}-\frac{1}{z+a}\right]

=K q \frac{2a}{z^2-a^2}

\because r

for the point is

\sqrt{x^2+y^2+a^2}

\therefore V_{(x, y, 0)} = \frac{K q}{(x^2+y^2+a^2)^{1/2}}

+ \frac{K(-q)}{(x^2+y^2+a^2)^{1/2}}

=0.

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