The equipotential surfaces corresponding to bsingle positive charge are concentric spherical shells with the charge at its origin. The spacing between the surfaces for the same change in potential

Question

The equipotential surfaces corresponding to bsingle positive charge are concentric spherical shells with the charge at its origin. The spacing between the surfaces for the same change in potential (A) is uniform throughout the field (B) is getting closer as r arrow ∞ (C) is getting closer as arrow 0 (D) can be.varied as one wishes to

Tardigrade Admin · Accepted Answer

Let the equipotential surfaces have potential

V_{1}, V_{2}, V_{3}

with,

V_{1} > V_{2} > V_{3}

Also, let

V_{3} - V_{2} = V_{2} - V_{1} = R

Potential of a charge is given by

k Q / r

V = k Q / r \Rightarrow r = k Q / V

Distance between the equipotential surfaces is

k Q (\frac{1}{V _{2}} - \frac{1}{V _{3}}) and k Q (\frac{1}{V _{1}} - \frac{1}{V _{2}})

or

k Q (\frac{V _{3} - V _{2}}{V _{3} V _{2}})

and

k Q (\frac{V _{2} - V _{1}}{V _{2} V _{1}})

or

k Q (\frac{R}{V _{3} V _{2}})

and

k Q (\frac{R}{V _{2} V _{1}})

Clearly

V_{2} V_{1} > V_{3} V_{2}

Thus

k Q (\frac{R}{V _{3} V _{2}}) > k Q (\frac{R}{V _{2} V _{1}})

Therefore spacing between surfaces decreases as potential increases i.e.

r

decreases.

Q. The equipotential surfaces corresponding to bsingle positive charge are concentric spherical shells with the charge at its origin. The spacing between the surfaces for the same change in potential

is uniform throughout the field

is getting closer as r $\to \infty$

is getting closer as $\to 0$

can be.varied as one wishes to

Q. The equipotential surfaces corresponding to bsingle positive charge are concentric spherical shells with the charge at its origin. The spacing between the surfaces for the same change in potential

is uniform throughout the field

is getting closer as r →∞

is getting closer as →0

can be.varied as one wishes to

is getting closer as r $\to \infty$

is getting closer as $\to 0$