The electric potential at a point in free spa | vedclass

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Question

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

Vedclass · Accepted Answer

$4\pi \varepsilon_0 Q 	imes 10^{22} \, V/m$

Vedclass · Answer

(C) The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r}$.Given $V = Q 	imes 10^{11} \, V$,we have $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} = Q 	imes 10^{11}$.This simplifies to $\frac{1}{r} = 4 \pi \varepsilon_0 	imes 10^{11}$.The electric field $E$ at that point is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2} = \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} ight) \cdot \frac{1}{r}$.Substituting the values,$E = (Q 	imes 10^{11}) 	imes (4 \pi \varepsilon_0 	imes 10^{11})$.Therefore,$E = 4 \pi \varepsilon_0 Q 	imes 10^{22} \, V/m$.

The electric potential at a point in free space due to charge $Q$ coulomb is $V = Q \times 10^{11} \, V$. The electric field at that point is:

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