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A free ball of 10 \mathrm{~g} has a positive charge of 6 \times 10^{-7} \mathrm{C}. Define the nature and magnitude of charge that must be given to a second ball placed 7 \mathrm{~cm} vertically below the former free ball so as to obtain a stationary position for the upper free ball?

Option: 1

8.89 \times 10^{-8} \mathrm{C}\, \, (positive)


Option: 2

8.89 \times 10^{-8} \mathrm{C} \, \, (negative)


Option: 3

8.89 \times 10^{-7} \mathrm{C}\, \, (negative)


Option: 4

8.89 \times 10^{-7} \mathrm{C}\, \, (positive)


Answers (1)

best_answer

Let \mathrm{Q}  be the charge on the second ball, A stationary position for the first ball can only be obtained if charges on both balls are of the same sign. To keep the first ball stationary, the net force acting on it should cancel out to be zero.
The force of gravity acting downward =\mathrm{Mg}
The weight of the upper ball is balanced by the electrostatic repulsion between the balls. For first ball to remain stationary,

F=m_1 g=\frac{Q q}{4 \pi \epsilon_o(0.07)^2}

Where \mathrm{q} is the charge on second ball and 0.07 \mathrm{~m} is the distance between two balls provided. 

After putting the correct values and doing all the calculations, answer will turn out to be:

8.89 \times 10^{-8} C \text { (positive) }

Posted by

Ritika Kankaria

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