MCQ
A solid sphere of radius $R$ is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre?
- A$R$
- B$R/2$
- ✓$R/3$
- D$2R$
✓
Answer
Correct option: C.
$R/3$
c
Potential at the center of sphere is $V_{c}=\frac{3}{2} \frac{k Q}{R}$
Potential at the center of sphere is $V_{c}=\frac{3}{2} \frac{k Q}{R}$
Potential at some distance $\times$ from center $=\frac{K Q}{x}$
$\frac{k Q}{x}=\frac{1}{2}\left(\frac{3}{2} \frac{k Q}{R}\right)$ or $x=4 R / 3$
Distance from surface is $x-R=R / 3$
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