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Electrostatics Part 1

Section: Physics

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Question : 90 of 100

Marks: +1, -0

Two thin wire rings each having a radius

R

are placed at a distance

d

apart with their axes coinciding. The charges on the two rings are

+q

-q

. The potential difference between the centres of the two rings is

$\;\frac{q}{2\pi E_0} \left[ \;\frac{1}{R} - \;\frac{1}{\sqrt{R^2+d^2}} \right]$
$\;\frac{qR}{4\pi\in_0 d^2}$
$\;\frac{q}{4\pi E_0} \left[ \;\frac{1}{R} - \;\frac{1}{\sqrt{R^2+d^2}} \right]$
zero

Solution:

V_A=V_{\;\text{self }\;}+V_{due}

to

(2)

\Rightarrow V_A= \;\frac{1}{4\pi\varepsilon_0} \left[ \;\frac{q}{R} - \;\frac{q}{\sqrt{R^2+d^2}} \right]

V_B=V_{self}+V_{due}

to

(1)

\Rightarrow V_B= \;\frac{1}{4\pi\varepsilon_0} \left[ \;\frac{-q}{R} + \;\frac{q}{\sqrt{R^2+d^2}} \right]

\Delta V=V_A-V_B

= \;\frac{1}{4\pi\varepsilon_0} \left[ \;\frac{q}{R} + \;\frac{q}{R} - \;\frac{q}{\sqrt{R^2+d^2}} - \;\frac{q}{\sqrt{R^2+d^2}} \right]

= \;\frac{1}{2\pi\varepsilon_0} \left[ \;\frac{q}{R} - \;\frac{q}{\sqrt{R^2+d^2}} \right]

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