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  2. Solve! - Electrostatics - JEE Main-6
# Engineering

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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