A pendulum bob of mass 80 mg and carrying a charge of 2 × 10^{−8} C is at rest in a uniform, horizontal electric field of 20 kVm^{−1}. Find the tension in the thread.

#### Solution

Given:

Magnitude of the charge, q = 2.0 × 10^{−8} C

Mass of the bob, m = 80 mg = 80 × 10^{−6} kg

Electric field, E = 20 kVm^{−1} = 20 × 10^{3} Vm^{−1}

Let the direction of the electric field be from the left to right. Tension in the string is T.

From the figure,

Tcosθ = mg ...(1)

Tsinθ = qE ...(2)

\[\Rightarrow \tan\theta = \frac{qE}{mg}\]

\[ = \frac{20 \times {10}^3 \times 2 \times {10}^{- 8}}{80 \times {10}^{- 6} \times 9 . 8}\]

\[\Rightarrow \tan\theta = \frac{5}{9 . 8}\]

\[ \Rightarrow \theta = 27^\circ\]

From equation (1),

\[T = \frac{mg}{cos\theta} = \frac{80 \times {10}^{- 6} \times 9 . 8}{\cos27^\circ}\]

\[ \Rightarrow T = 8 . 8 \times {10}^{- 4} \]N