A particle having a charge of $2.0 \times 10^{- 4} C$ is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. Teh mass of the bob be given so that the string becomes loose?

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Solution

Mass of the bob = 100g = 0.1 g
So, tension in th estring
$= 0.1 \times 9.8 = 0.98 N$
For the tension to be 0, the charge below should reel the 1st bob
and F = mg
$\Rightarrow \frac{9 \times 10^{9} \times q \times 2 \times 10^{- 4}}{10^{- 2}} = 10^{- 1} \times 10 \Rightarrow q = 5.4 \times 10^{- 9} C$

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A particle having a charge of 2.0×10−4C is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. Teh mass of the bob be given so that the string becomes loose?

Mass of the bob = 100g = 0.1 g So, tension in th estring =0.1×9.8=0.98N For the tension to be 0, the charge below should reel the 1st bob and F = mg ⇒9×109×q×2×10−410−2=10−1×10⇒q=5.4×10−9C

A particle having a charge of $2.0 \times 10^{- 4} C$ is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. Teh mass of the bob be given so that the string becomes loose?