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The bob of a simple pendulum has mass 2 g and charge of $5.0\; \mu C$ . It is at rest in a uniform horizontal electric field of intensity $2000\; V/m$ . At equilibrium, the angle that the pendulum makes with the vertical is :

(take $g=10\; m/s^{2}$ )

Option 1)
$\tan ^{-1}(5.0)$

Option 2)
$\tan ^{-1}(2.0)$

Option 3)
$\tan ^{-1}(0.2)$

Option 4)
$\tan ^{-1}(0.5)$

Answers (1)

S solutionqc

$m=2\: g=2\times 10^{-3}kg$

$9 = 5\:\mu C$

$E = 2000\; V/m$

$\tan \Theta =\frac{qE}{mg}=\frac{5\times 10^{-6}\times 2000}{2\times 10^{-3}\times 10}=\frac{1}{2}$

$\Theta =\tan ^{-1}(0.5)$

Option 1)

$\tan ^{-1}(5.0)$

Option 2)

$\tan ^{-1}(2.0)$

Option 3)

$\tan ^{-1}(0.2)$

Option 4)

$\tan ^{-1}(0.5)$

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