Question

The electric field due to a charge at a distance of 3m from it is 500n/c the magnitude of the charge is

Answer 1

your answer is in attachment

Answer 2

The magnitude of electric charge is $5\times 10^{-7}\ C$.

Explanation:

It is given that,

Electric field, E = 500 N/C

Distance, d = 3 m

To find,

Magnitude of charge

Solution.

The electric field at a distance d from the charge is given by :

$E=\dfrac{kq}{d^2}$

$q=\dfrac{Ed^2}{k}$

$q=\dfrac{500\times (3)^2}{9\times 10^9}$

$q=5\times 10^{-7}\ C$

So, the magnitude of electric charge is $5\times 10^{-7}\ C$.

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Electric field

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