Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle θ from the vertical. If the mass of each charge is m, then the electrostatic potential at the centre of line joining them will be ((1/4π ∈0)=k).

Question

Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle θ from the vertical. If the mass of each charge is m, then the electrostatic potential at the centre of line joining them will be ((1/4π ∈0)=k). (A) 2√k mg tan θ (B) √k mg tan θ (C) 4√k mg/tan θ (D) √mg/tan θ

Tardigrade Admin · Accepted Answer

In equilibrium, the expressions are given as,

F = T sin θ

(1)

m g = T cos θ

(2)
From above equations, it can be written as,

tan θ = \frac{F}{m g}

The electric force is given as,

F = \frac{q ^{2}}{4 π ε _{0} x ^{2}}

Substitute the value of F in equation (3), we get

tan θ = \frac{q ^{2}}{4 π ε _{0} m g x ^{2}}

x = \frac{q ^{2}}{4 π ε _{0} m g t a n θ}

The electric potential is given as,

V = \frac{k q}{\frac{x}{2}} + \frac{k q}{\frac{x}{2}}

V = \frac{4 k q}{x}

V = 4 km g tan θ

Thus, the electric potential at the centre of the line is

4 km g tan θ

.

$2 k m g t an θ$

$k m g t an θ$

$4 k m g / t an θ$

$m g / t an θ$

2kmgtanθ​

kmgtanθ​

4kmg/tanθ​

mg/tanθ​

$2 k m g t an θ$

$k m g t an θ$

$4 k m g / t an θ$

$m g / t an θ$