Question

A charged particle of mass 1.0 g is suspended through a silk thread of length 40 cm in a horizontal electric field of intensity $4.0\times {10}^{4}N{C}^{-1}$ as shown in the figure. If the particle stays in equilibrium at a distance of 24 cm from the wall, find the net charge on the particle?

• A. q = 1.8 × 10^–7 C
• B. q = 3.2× 10^–7 C
• C. q = –3.2× 10^–7 C
• D. None of these

Answer 1

The correct option is A

q = 1.8 × 10^–7 C

Let’s draw the free body diagram of the charged particle-

The situation is shown in figure. The forces acting on the particle are

(i) The electric force F = qE horizontally leftwards,

(ii) The force of gravity mg downward and

(iii) The tension T along the thread

As the particle is at rest, these forces should add to zero.

T cos θ = mg and T sin θ = F

or, F = mg tan θ

From the figure,

$Sin\theta =\frac{24}{40}=\frac{3}{5}$

$Thus,tan\theta =\frac{3}{4}$ From (i),

$q(4.0\times {10}^{4}N{C}^{-1})=(1.0\times {10}^{-3}kg)(9.8m-s^{-2})\left(\frac{3}{4}\right)$

Giving $q=1.8\times {10}^{-7}C$