Question

Two equal charges are separated by a distance of 2 m in free space.Calculate the magnitude of charges so that force between them is equal to weight of a 20 kg child.

Answer 1

Answer:

kq^2/d^2 = mg

q^2 = mg d^2 /k

q = √(mg d^2/k)

q = d √(mg/k)

Put the values for answer

d = 2

mg = 20×10 = 200

k = 9 x 10^9

Answer 2

The magnitude of charges is $2.95\times 10^{-4}\ C$.

Explanation:

Let charges are q. These two charges are separated by a distance of 2 m in free space. The force between them is equal to weight of a 20 kg child.

The electric force is given by :

$F=\dfrac{kq^2}{d^2}\\\\mg=\dfrac{kq^2}{d^2}\\\\q=\sqrt{\dfrac{mgd^2}{k}} \\\\q=\sqrt{\dfrac{20\times 9.8\times (2)^2}{9\times 10^9}} \\\\q=2.95\times 10^{-4}\ C$

So, the magnitude of charges is $2.95\times 10^{-4}\ C$.

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Electric force

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