NEET 2024 — Physics solution

NEET2024Physics

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 $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9}$ SI units)

A
$0.5 \times 1 0^{5}$
B
zero
C
$3 \times 1 0^{5}$
D
$1 \times 1 0^{5}$

Solution

Inside (and on the surface of) a thin uniformly-charged spherical shell, the electric potential is constant and equal to the surface value.

Hence $V_{C} = V_{P}$ and $V_{C} - V_{P} = 0$ .

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