A) \[\frac{v}{6r}\]
B) \[\frac{v}{4r}\]
C) \[\frac{V}{3r}\]
D) \[\frac{V}{2r}\]
Correct Answer: A
Solution :
The potential at a point distant r from charge q is \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}\] At a distance of 3r \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{3r}\] Potential difference \[\Delta V={{V}_{1}}-{{V}_{2}}\] \[\Delta V=\frac{q}{4\pi {{\varepsilon }_{0}}r}-\frac{q}{4\pi {{\varepsilon }_{0}}3r}=\frac{2q}{4\pi {{\varepsilon }_{0}}3r}\] ?. (i) Electric field intensity \[E=\frac{q}{4\pi {{\varepsilon }_{0}}}.\frac{1}{{{(3r)}^{2}}}\] ...(ii) From Eqs. (i) and (ii), we get \[\frac{E}{V}=\frac{q}{4\pi {{\varepsilon }_{0}}\times q{{r}^{2}}}\times \frac{4\pi {{\varepsilon }_{0}}\times 3r}{2q}\] \[\Rightarrow \] \[\frac{E}{V}=\frac{1}{6r}\] \[\Rightarrow \] \[E=\frac{V}{6r}\]
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