Electric potential at a point due to a point charge of is . The distance of from the point charge is (Assume, ) [2023]

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Solution

Q.
Electric potential at a point $' P'$ due to a point charge of $5 \times 10^{- 9} C$ is $50 V$ . The distance of $' P'$ from the point charge is

(Assume, $\frac{1}{4 π ε_{0}} = 9 \times 10^{- 9} {Nm}^{2} C^{- 2}$ ) [2023]

1 3 cm

2 90 cm

3 9 cm

4 0.9 cm

Ans.
(2)

$V_{P} = \frac{K Q}{r}$

$50 = \frac{9 \times 10^{9} \times 5 \times 10^{- 9}}{r}$

$r = \frac{45}{50} = \frac{9}{10} = 0.9 m = 90 cm$

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