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Two identical thin rings each of radius R, are coaxially placed a distance R apart. If Q 1 and Q 2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is

  • Option 1)

    Zero

  • Option 2)

    \frac{q\left ( Q_{1}-Q_{2}\left ( \sqrt{2} -1\right ) \right )}{4\pi \varepsilon _{0}R\sqrt{2}}

     

     

     

  • Option 3)

    \frac{q\left ( Q_{1}+Q_{2} \right )\sqrt{2}}{4\pi \varepsilon _{0}R}

  • Option 4)

    \frac{q\left ( \frac{Q_{1}}{Q_{2}} \right )\left ( \sqrt{2}-1 \right )}{4\pi \varepsilon _{0}R\sqrt{2}}

 

Answers (2)

best_answer

 

As we learned

E and V at a point P that lies on the axis of ring -

\dpi{100} E_{x}=\frac{kQx}{\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}  ,   V=\frac{kQ}{\left ( x^{2}+R^{2} \right )^{\frac{1}{2}}}

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Charged Circular ring -

A charged circular ring of radius R and charge Q.

- wherein

 

 

 

 

Potential at the centre of first ring V_{A}=\frac{Q_{1}}{4\pi \varepsilon _{0}R}+\frac{Q_{2}}{4\pi \varepsilon _{0}\sqrt{R^{2}+R^{2}}}

Potential at the centre of second ring V_{B}=\frac{Q_{2}}{4\pi \varepsilon _{0}R}+\frac{Q_{1}}{4\pi \varepsilon _{0}\sqrt{R^{2}+R^{2}}}

Potential difference between the two centres V_{A}-V_{B}=\frac{\left ( \sqrt{2}-1 \right )\left ( Q_{1}-Q_{2} \right )}{4\pi \varepsilon _{0}R\sqrt{2}}

\therefore Work\; done     W=\frac{q\left ( \sqrt{2}-1 \right )\left ( Q_{1}-Q_{2} \right )}{4\pi \varepsilon _{0}R\sqrt{2}}

 

 


Option 1)

Zero

Option 2)

\frac{q\left ( Q_{1}-Q_{2}\left ( \sqrt{2} -1\right ) \right )}{4\pi \varepsilon _{0}R\sqrt{2}}

 

 

 

Option 3)

\frac{q\left ( Q_{1}+Q_{2} \right )\sqrt{2}}{4\pi \varepsilon _{0}R}

Option 4)

\frac{q\left ( \frac{Q_{1}}{Q_{2}} \right )\left ( \sqrt{2}-1 \right )}{4\pi \varepsilon _{0}R\sqrt{2}}

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