The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11}…

The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11} \mathrm{~V}$. The electric field at that point is
  1. $4 \pi \varepsilon_0 Q \times 10^{22} \mathrm{~V} / \mathrm{m}$
  2. $12 \pi \varepsilon_0 \mathrm{Q} \times 10^{20} \mathrm{~V} / \mathrm{m}$
  3. $4 \pi \varepsilon_0 Q \times 10^{20} \mathrm{~V} / \mathrm{m}$
  4. $12 \pi \varepsilon_0 Q \times 10^{22} \mathrm{~V} / \mathrm{m}$

Solution

Key Idea : Search for the relations of electric potential and electric field at a particular point. At any point, electric potential due to charge $Q$ is $V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r}$ where $r$ is the distance of observation point from the charge. At the same point, electric field is $E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$ Combining Eqs. (i) and (ii), we have $\begin{aligned} E & =\frac{4 \pi \varepsilon_0 V^2}{Q}=\frac{4 \pi \varepsilon_0 \times\left(Q \times 10^{11}\right)^2}{Q} \\ & =4 \pi \varepsilon_0 \mathrm{Q} \times 10^{22} \mathrm{~V} / \mathrm{m} \end{aligned}$

Asked in: NEET 2008 (Screening)

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