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, ± z)

and

(x, y, 0)

Solution:

(a)

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

\;=K q [\;{1}/{z-a}-\;{1}/{z+a}] =K q \;{2 a}/{z^2-a^2}

Potential at pt.

(0,0-z)

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

\;=K q [\;{1}/{z-a}-\;{1}/{z+a}]

\;=K q \;{2 a}/{z^2-a^2}

∵ r

for the point is

{√{x^2+y^2+a^2}}

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

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

\;=0 .

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