Two thin wire rings each having a radius r

Asked by admin @ 13/08/2021 in Physics viewed by 327 People

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.

Answered by admin @ 13/08/2021

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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