There are two thin wire rings,each of radius | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the centres of the rings be $A$ and $B$. The distance between them is $d = \sqrt{3} R$.Potential at centre $A$ due to ring $1$ (charge $q$) is

Vedclass · Accepted Answer

$\frac{q}{4 \pi \varepsilon_0 R}$

Vedclass · Answer

(B) Let the centres of the rings be $A$ and $B$. The distance between them is $d = \sqrt{3} R$.Potential at centre $A$ due to ring $1$ (charge $q$) is $V_{A1} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{R}$.Potential at centre $A$ due to ring $2$ (charge $-q$) is $V_{A2} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + 3R^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{2R}$.Total potential at $A$ is $V_A = V_{A1} + V_{A2} = \frac{1}{4 \pi \varepsilon_0} (\frac{q}{R} - \frac{q}{2R}) = \frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.Similarly,potential at centre $B$ due to ring $2$ (charge $-q$) is $V_{B2} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{R}$.Potential at centre $B$ due to ring $1$ (charge $q$) is $V_{B1} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.Total potential at $B$ is $V_B = V_{B1} + V_{B2} = \frac{1}{4 \pi \varepsilon_0} (\frac{q}{2R} - \frac{q}{R}) = -\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.The magnitude of potential difference is $|V_A - V_B| = |\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R} - (-\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R})| = |\frac{1}{4 \pi \varepsilon_0} \frac{2q}{2R}| = \frac{q}{4 \pi \varepsilon_0 R}$.

There are two thin wire rings,each of radius $R$,whose axes coincide. The charges of the rings are $q$ and $-q$. The magnitude of potential difference between the centres of the rings separated by a distance $\sqrt{3} R$ is

Similar Questions

What is the electric potential at point $A$ due to the given charges?

$A$ conductor with a positive charge:

Calculate the work done in taking a charge $q_0 = -2 \times 10^{-9} \, C$ from point $A$ to point $B$ via point $C$,as shown in the diagram. The source charge $q = 8 \, mC$ is located at the origin $O$. (in $, J$)

Two spherical conductors of radii $4 \ cm$ and $5 \ cm$ are charged to the same potential. If $\sigma_1$ and $\sigma_2$ are the respective values of the surface charge density on the two conductors,then the ratio $\sigma_1 : \sigma_2$ is

An electrostatic paint sprayer has a metal sphere of diameter $18 \, cm$ and at a potential of $25 \, kV$. The charge on the metal sphere is: (in $ \, \mu C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module