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  4. A force of 2.25 N acts on a charge of 15 × 10-4 C. The intensity of electric field at that point is

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Q. A force of $2.25\,N$ acts on a charge of $15 \times 10^{-4}\, C$. The intensity of electric field at that point is

Electric Charges and Fields

Solution:

Electric field, $E=\frac{F}{q}$
$=\frac{2.25\,N}{15\times10^{-4}\,C}$
$=1500\,N\,C^{-1}$