A) \[\frac{V}{2r}\]
B) \[\frac{V}{3r}\]
C) \[\frac{V}{4r}\]
D) \[\frac{V}{6r}\]
Correct Answer: D
Solution :
The potential at a distance r, due to charge q is \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}\] Potential at a distance (3r) is \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{3r}\] Difference in potential \[=\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{r}-\frac{1}{3r} \right]\] \[\Rightarrow \] \[V=\frac{2q}{4\pi {{\varepsilon }_{0}}\times 3r}\] Intensity of electric field \[E=\frac{q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{(3r)}^{2}}}\] \[\therefore \] \[\frac{E}{V}=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\times \frac{4\pi {{\varepsilon }_{0}}3r}{2q}\] \[\Rightarrow \] \[\frac{E}{V}=\frac{1}{6r}\] \[\Rightarrow \] \[E=\frac{V}{6r}\]You need to login to perform this action.
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