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Question: The electric potential at a point in free space due to a charge Q coulomb is \(Q \times 10^{11}\)V. ...

The electric potential at a point in free space due to a charge Q coulomb is Q×1011Q \times 10^{11}V. The electric field at that point is.

A

12πε0Q×1022Vm−112\pi\varepsilon_{0}Q \times 10^{22}Vm^{- 1}

B

4πε0Q×1022Vm−14\pi\varepsilon_{0}Q \times 10^{22}Vm^{- 1}

C

12πε0Q×1020Vm−112\pi\varepsilon_{0}Q \times 10^{20}Vm^{- 1}

D

4πε0Q×1020Vm−14\pi\varepsilon_{0}Q \times 10^{20}Vm^{- 1}

Answer

4πε0Q×1022Vm−14\pi\varepsilon_{0}Q \times 10^{22}Vm^{- 1}

Explanation

Solution

: Here, V=Q4πε0r=Q×1011V = \frac{Q}{4\pi\varepsilon_{0}r} = Q \times 10^{11}

∴4πε0r=10−11\therefore 4\pi\varepsilon_{0}r = 10^{- 11}

Now, E=Q4πε0r2=Q×4πε0(4πε0r)2=4πε0Q(10−11)E = \frac{Q}{4\pi\varepsilon_{0}r^{2}} = \frac{Q \times 4\pi\varepsilon_{0}}{(4\pi\varepsilon_{0}r)^{2}} = \frac{4\pi\varepsilon_{0}Q}{(10^{- 11})} (Using (i))

=4πε0Q×1022Vm−1= 4\pi\varepsilon_{0}Q \times 10^{22}Vm^{- 1}