$$E = -\frac{dV}{dr} = -\frac{d}{dr}\left(\frac{q}{4\pi\varepsilon_0 r}\right) = \frac{q}{4\pi\varepsilon_0 r^2}$$
Back to NEET UG 2025 Electrostatics
NEET UG 2025 — Physics Electrostatics
easy
mcq
2025
Official previous-year question
Verified 30 May 2026.
Question
The electric potential at distance $r$ from a point charge $q$ is $V = \frac{q}{4\pi\varepsilon_0 r}$. The electric field $E$ is:
Options
- A
$E = -\frac{dV}{dr} = \frac{q}{4\pi\varepsilon_0 r^2}$
- B
$E = \frac{dV}{dr} = -\frac{q}{4\pi\varepsilon_0 r^2}$
- C
$E = \frac{q}{4\pi\varepsilon_0 r}$
- D
$E = \frac{q^2}{4\pi\varepsilon_0 r^2}$
Solution
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