If a conducting sphere of radius R is charged. Then the electric field at a distance r (r R) from the centre

Question

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):

Accepted Answer

1. **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 $Q = V \cdot R \cdot (4\pi\varepsilon_0)$.
2. **Electric Field Outside:** The electric field ($E$) at a distance $r$ (where $r > R$) outside the sphere is given by Coulomb's law: $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$.
3. **Substitution:** Substitute the expression for $Q$ from step 1 into the electric field equation:
   $E = \frac{1}{4\pi\varepsilon_0} \frac{(V \cdot R \cdot 4\pi\varepsilon_0)}{r^2}$
   $E = \frac{VR}{r^2}$.
4. **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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