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  4. The electric potential and electric field at a point due to a point charge are 1800 V and 900 N / C respectively. Then magnitude of the point charge should be :-

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Q. The electric potential and electric field at a point due to a point charge are $1800\, V$ and $900\, N / C$ respectively. Then magnitude of the point charge should be :-

Electric Charges and Fields

Solution:

$V =\frac{ KQ }{ r } $
$E =\frac{ KQ }{ r ^{2}}$
$\frac{ E }{ V }=\frac{1}{ r } $
$\Rightarrow \frac{900}{1800}=\frac{1}{ r }$
$r =2\, m$
$1800=\frac{9 \times 10^{9} \times Q }{2} $
$\Rightarrow Q =0.4 \,\mu C$