Question

Consider a uniformly charged wire that has the
form of a circular loop with radius b. Consider
two points on the axis of the loop. P, is at a
distance b from the loop's center, and P, is at a
distance 2b from the loop's center. The potential
V is zero, very far away from the loop. At P,
and P, the potentials are V, and V, respectively.
What is V, in terms of V,?
(A) 2V
(B)
2b
(C)
Jiu
Lovin
fdipole moment​

Answer 1

Given:

A uniformly charged wire is in the form of a ring of radius "b". Point P1 is on the axis of the ring at a distance of from the loop's centre. Point P2 is at a distance 2b .

To find:

Express the potential at P1 and P2 in terms of each other.

Calculation:

At point P1 , let potential be V1.

$\therefore \: dV_{1} = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{dq}{ \sqrt{ {b}^{2} + {b}^{2} } } \bigg \}$

Integrating on both sides :

$\displaystyle = > \int \: dV_{1} = \dfrac{1}{4\pi\epsilon_{0}} \int\dfrac{dq}{ \sqrt{ {b}^{2} + {b}^{2} } }$

$\displaystyle = > V_{1} = \dfrac{1}{4\pi\epsilon_{0} \sqrt{2 {b}^{2} } } \int \: dq$

$\displaystyle = > V_{1} = \dfrac{q}{4\pi\epsilon_{0}b \sqrt{2 } } \: \: \: \: .......(1)$

Similarly, at Point P2 , let potential be V2.

$\therefore \: dV_{2} = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{dq}{ \sqrt{ {b}^{2} + {(2b)}^{2} } } \bigg \}$

Integrating on both sides:

$\displaystyle = > \int \: dV_{2} = \dfrac{1}{4\pi\epsilon_{0}} \int\dfrac{dq}{ \sqrt{5 {b}^{2}} }$

$\displaystyle = > V_{2} = \dfrac{1}{4\pi\epsilon_{0} \sqrt{5{b}^{2} } } \int \: dq$

$\displaystyle = > V_{2} = \dfrac{q}{4\pi\epsilon_{0}b \sqrt{5 } } \: \: \: \: .......(2)$

Comparing between eq.(1) and eq.(2) , we get:

$\therefore \: \: \dfrac{V_{1}}{V_{2}} = \dfrac{ \bigg(\dfrac{q}{4\pi\epsilon_{0}b \sqrt{2 } } \bigg) }{ \bigg(\dfrac{q}{4\pi\epsilon_{0}b \sqrt{5 } } \bigg)}$

$= > \: \: \dfrac{V_{1}}{V_{2}} = \dfrac{ \sqrt{5} }{ \sqrt{2} }$

$= > \: \: V_{1} = \bigg(\dfrac{ \sqrt{5} }{ \sqrt{2} } \bigg) \: V_{2}$

So final answer is :

$\boxed{ \red{ \bold{ \: \: V_{1} = \bigg(\dfrac{ \sqrt{5} }{ \sqrt{2} } \bigg) \: V_{2}}}}$