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# The electric potential at a point in space due to charge $Q$ coulomb is $Q \times {10^{11}}V$. The value of electric field due to charge $Q$ at that point is equal to:(A) $12\pi { \in _0}Q \times {10^{22}}V{m^{ - 1}}$(B) $4\pi { \in _0}Q \times {10^{22}}V{m^{ - 1}}$(C) $12\pi { \in _0}Q \times {10^{20}}V{m^{ - 1}}$(D) $4\pi { \in _0}Q \times {10^{20}}V{m^{ - 1}}$

Last updated date: 16th Jul 2024
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Hint Electric potential at a point in space is directly proportional to charge of source charge and inversely proportional to distance between source charge and that point. An electric field at a point is directly proportional to the charge of source charge and inversely proportional to square of distance between the source charge and that point. Use this relation between electric potential and electric field to find electric field using electric potential at that point.

Complete step by step solution
Electric potential at a point in space is directly proportional to charge of source charge and inversely proportional to distance between source charge and that point.
$V = \dfrac{1}{{4\pi { \in _0}}}\dfrac{Q}{R}$ (1)
Where $R$ is distance between source charge $Q$ and the given point.
an electric field at a point is directly proportional to the charge of source charge and inversely proportional to square of distance between the source charge and that point.
$E = \dfrac{1}{{4\pi { \in _0}}}\dfrac{Q}{{{R^2}}}$ (2)
Combining equation (1) and (2), we have
$E = \dfrac{{4\pi { \in _0}{V^2}}}{Q}$
Substituting $V = Q \times {10^{11}}$ as given in question, we get
$E = \dfrac{{4\pi { \in _0}{{(Q \times {{10}^{11}})}^2}}}{Q} = 4\pi { \in _0}Q \times {10^{22}}V{m^{ - 1}}$

Hence, the correct answer is option B.

Note Electric field at a point in space due to charge $Q$ gives the value of force applied on unit positive charge placed at that point, where electric potential is potential energy of a unit positive charge at that point due to charge $Q$.