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Two thin wire rings each having a radius R are placed at a distance apart with their axes coinciding. The charges on the two rings are +Q and -Q  The potential difference between the centers of the two rings is

  • Option 1)

    zero

  • Option 2)

    \frac{Q}{4\pi \varepsilon _{0}}\left [ \frac{1}{R} -\frac{1}{\sqrt{R^{2}+d^{2}}}\right ]

  • Option 3)

    \frac{QR}{4\pi \varepsilon _{0}d^{2}}

  • Option 4)

    \frac{Q}{2\pi \varepsilon _{0}}\left [ \frac{1}{R} -\frac{1}{\sqrt{R^{2}+d^{2}}}\right ]

 

Answers (2)

As we learnt in

Electric Potential -

V=\frac{w}{q_{0}}

- wherein

w - work done 

q0 - unit charge.

 

 

V_{A}= \frac{1}{4\pi \varepsilon _{0}}\frac{Q}{R}-\frac{1}{4\pi \varepsilon _{0}}\frac{Q}{\sqrt{R^{2}+d^{2}}}

V_{B}= \frac{1}{4\pi \varepsilon _{0}}\frac{\left ( -Q \right )}{R}+\frac{1}{4\pi \varepsilon _{0}}\frac{Q}{\sqrt{R^{2}+d^{2}}}

\therefore \: \: V_{A}-V_{B}= \frac{1\times Q}{4\pi \varepsilon _{0}}\left [ \frac{2}{R} -\frac{2}{\sqrt{R^{2}+d^{2}}}\right ]

\: \: \: \: = \frac{ Q}{2\pi \varepsilon _{0}}\left [ \frac{1}{R} -\frac{1}{\sqrt{R^{2}+d^{2}}}\right ]

 


Option 1)

zero

Incorrect

Option 2)

\frac{Q}{4\pi \varepsilon _{0}}\left [ \frac{1}{R} -\frac{1}{\sqrt{R^{2}+d^{2}}}\right ]

Incorrect

Option 3)

\frac{QR}{4\pi \varepsilon _{0}d^{2}}

Incorrect

Option 4)

\frac{Q}{2\pi \varepsilon _{0}}\left [ \frac{1}{R} -\frac{1}{\sqrt{R^{2}+d^{2}}}\right ]

Correct

Posted by

Sabhrant Ambastha

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