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(C) The formula for electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$.Giv

Vedclass · Accepted Answer

$90$

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(C) The formula for electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$.Given: $V = 50 \; V$,$Q = 5 	imes 10^{-9} \; C$,and $k = \frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 \; N m^2 C^{-2}$.Substituting the values into the formula: $50 = \frac{9 	imes 10^9 	imes 5 	imes 10^{-9}}{r}$.Simplifying the numerator: $50 = \frac{45}{r}$.Solving for $r$: $r = \frac{45}{50} = 0.9 \; m$.Converting meters to centimeters: $r = 0.9 	imes 100 \; cm = 90 \; cm$.

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 ......... $cm$. (Assume,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \; N m^2 C^{-2}$)

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