Question

The electric potential at a certain distance from a source charge is 600 V and electric field strength at that point is 150 N/C. What is the distance of observation point from the source charge ?

Easy

Solution

$E = \frac{1}{4 {πε}_{0}} \times \frac{Q}{r^{2}} = 150 N / C and V = \frac{1}{4 {πε}_{0}} \times \frac{Q}{r} = 600 V$
Dividing, we get r = V/E = 600/150 = 4m

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