A force of 40 N acts on a charge of 0.25 C in a region of space. Find out the electric field at the point of the charge ?

10 N/C

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100 N/C

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160 N/C

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40 N/C

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0.00625 N/C

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Solution

The correct option is D 160 N/C
Given : $F = 40$ N $q = 0.25$ C
Thus electric field at that point $E = \frac{F}{q} = \frac{40}{0.25} = 160$ N/C

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