A particle a having a charge of `5.0 xx 10 ^(-7) C` is fixed in. a vertical wall. A second particle B of mass 100 g and. having equal charge is supended by a silk thread. of length 30 cm form the wall. The point of suspension is. 30 cm above the particle A. Find the angle of the thread. with the vertical when it stays in equilibrium.

Explanation

To solve the problem step-by-step, we will analyze the forces acting on particle B and apply the principles of equilibrium and trigonometry.

### Step 1: Understand the Setup
We have two charged particles, A and B, both with a charge of \( q = 5.0 \times 10^{-7} \, C \). Particle A is fixed on a vertical wall, while particle B is suspended by a silk thread of length \( L = 30 \, cm = 0.3 \, m \) from a point 30 cm above particle A. The distance between the two charges is \( d = 60 \, cm = 0.6 \, m \) when B is in equilibrium.

### Step 2: Identify Forces Acting on Particle B
In equilibrium, the forces acting on particle B are:
1. The gravitational force \( F_g = mg \) acting downward.
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