Electrostatic Potential and Capacitance - Result Question 11
12. The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11}$ volts. The electric field at that point is
[2008]
(a) $4 \pi \varepsilon_0 Q \times 10^{22} volt / m$
(b) $12 \pi \varepsilon_0 Q \times 10^{20} volt / m$
(c) $4 \pi \varepsilon_0 Q \times 10^{20}$ volt $/ m$
(d) $12 \pi \varepsilon_0 Q \times 10^{22} volt / m$
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Answer:
Correct Answer: 12. (a)
Solution:
- (a) Given that, $V=Q \times 10^{11}$ volts
Electric potential at point is given by
$ \begin{aligned} & V=\frac{1}{4 \pi \in_0} \cdot \frac{Q}{r} \text{ or, } Q \times 10^{11}=\frac{1}{4 \pi \in_0} \cdot \frac{Q}{r} \\ & \Rightarrow r=\frac{1}{4 \pi \epsilon_0} \cdot 10^{-11} m . \\ & \text{ As }|E|=\frac{|V|}{r}=\frac{\frac{Q}{} \cdot 10^{11}}{4 \pi \epsilon_0} \cdot 10^{-11} \\ & =4 \pi \in_0 Q \times 10^{22} \text{ volt } m^{-1} . \end{aligned} $
