Physics, asked by TheMajesty27, 9 months ago

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 and -q. find the potential difference between the centres of the two rings.​

Answers

Answered by Anonymous
17

Answer:

\large\boxed{\sf{ \dfrac{1}{2\pi{\epsilon}_{0}}( \dfrac{q}{r}  -  \dfrac{q}{ \sqrt{ {r}^{2}  +  {d}^{2} } } )}}

Explanation:

Given that,

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 and -q.

\red{Note:-}See the attachment.

Now, to find the potential difference between the centres of the two rings.

Let the centres be A and B respectively.

Therefore, we will get,

V_{A}=V_{self}+V_{due\:to\:(2)}

 =  > V_{A} = \dfrac{1}{4\pi{\epsilon}_{0}}( \dfrac{q}{r}  -  \dfrac{q}{ \sqrt{ {r}^{2}  +  {d}^{2} } } )

Similarly, we have,

V_{B}=V_{self}+V_{due\:to\:(1)}

 =  > V_{B} = \dfrac{1}{4\pi{\epsilon}_{0}}( \dfrac{ - q}{r}  +  \dfrac{q}{ \sqrt{ {r}^{2}  +  {d}^{2} } } )

Therefore, we will get,

 =  >  \triangle V = V_{A} - V_{B} \\  \\  =  >   \triangle V  =  \dfrac{1}{4\pi{\epsilon}_{0}}( \dfrac{q}{r}  +  \dfrac{q}{r}  -  \dfrac{q}{  \sqrt{ {r}^{2}  +  {d}^{2}} }  -  \dfrac{q}{ \sqrt{ {r}^{2}  +  {d}^{2} } } ) \\  \\  =  >   \triangle V  =  \dfrac{1}{2\pi{\epsilon}_{0}}( \dfrac{q}{r}  -  \dfrac{q}{ \sqrt{ {r}^{2}  +  {d}^{2} } } )

Hence, the required potential difference is \bold{ \dfrac{1}{2\pi{\epsilon}_{0}}( \dfrac{q}{r}  -  \dfrac{q}{ \sqrt{ {r}^{2}  +  {d}^{2} } } )}

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