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2. Electric Potential and Capacitance
medium
A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$) shown in the figure is:
(Take $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$ $SI$ units)
A
$1 \times 10^5$
B
$0.5 \times 10^5$
C
Zero
D
$3 \times 10^5$
(NEET-2024)
Solution
For uniformly charged spherical shell,
$V=\frac{k q}{R} \quad(\text { For } r \leq R)$
$\therefore \quad V_c=V_p$
$V_c-V_p=\text { Zero }$
Standard 12
Physics
