A pendulum bob of mass $m = 71 mg$ , carrying a charge of $Q = 2 \times 10^{- 8} C$ , is at rest in a horizontal, uniform electric field $E = 20000 {Vm}^{- 1}$ . The tension $T$ in the thread of the pendulum is $T$ and the angle it makes with the vertical is $θ$ . Then

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$θ = 27 °$

$θ = 30 °$

$T = 803 μ N$

$T = 880 μ N$

answer is C, D.

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Detailed Solution

equating the forces in the vertical and horizontal directions

$T \cos θ = m g$ and $T \sin θ = F = Q E$

Dividing, $\tan θ = \frac{Q E}{m g} = \frac{2 \times 10^{- 8} \times 2 \times 10^{4}}{71 \times 10^{- 6} \times 9.8}$

$\tan θ = 0.57 ≅ \frac{1}{\sqrt{3}} θ = 30 ° l o w, T = \frac{71 \times 10^{- 6} \times 9.8}{\cos (30 °)} = 803 μ N$

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