Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

If a conducting sphere of radius R is charged. Then the electric field at a distance r (r > R) from the centre of the sphere would be, (V = potential on the surface of the sphere):

rV/R^2

R^2V/r^3

RV/r^2

V/r

Step-by-Step Solution

Potential on Surface: The electric potential ( $V$ ) on the surface of a charged conducting sphere of radius $R$ carrying charge $Q$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$ . From this, we can express the charge as $Q = V \cdot R \cdot (4\pi\varepsilon_0)$ .
Electric Field Outside: The electric field ( $E$ ) at a distance $r$ (where $r > R$ ) outside the sphere is given by Coulomb's law: $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$ .
Substitution: Substitute the expression for $Q$ from step 1 into the electric field equation: $E = \frac{1}{4\pi\varepsilon_0} \frac{(V \cdot R \cdot 4\pi\varepsilon_0)}{r^2}$ $E = \frac{VR}{r^2}$ .
Analogy: This relationship is mathematically analogous to the relationship between gravitational field and potential discussed in Class 11 Physics, Chapter 8.

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