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Question

(C) Given: Electric field $E = 500 \ NC^{-1}$,Electric potential $V = 30 \ V$.The relationship between electric field and potential for a point charge

Vedclass · Accepted Answer

$2 	imes 10^{-10} \ C$

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(C) Given: Electric field $E = 500 \ NC^{-1}$,Electric potential $V = 30 \ V$.The relationship between electric field and potential for a point charge is $E = \frac{V}{r}$.Therefore,the distance $r$ is $r = \frac{V}{E} = \frac{30}{500} = 0.06 \ m$.The formula for electric potential is $V = \frac{kq}{r}$,where $k = 9 	imes 10^9 \ Nm^2C^{-2}$.Rearranging for charge $q$: $q = \frac{V \cdot r}{k}$.Substituting the values: $q = \frac{30 	imes 0.06}{9 	imes 10^9} = \frac{1.8}{9 	imes 10^9} = 0.2 	imes 10^{-9} \ C$.Thus,$q = 2 	imes 10^{-10} \ C$.

The electric field and electric potential at a point due to a point charge are $500 \ NC^{-1}$ and $30 \ V$ respectively. The magnitude of the charge is:

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