Electric Field of Conducting Sphere – Rankers Physics

Electric Potential: Practice Problem & Solution

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)
\(\frac{RV}{r^2}\)
\(\frac{V}{r}\)
\(\frac{rV}{R^2}\)
\(\frac{R^2V}{r^3}\)

Solution Explained:

To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:

Potential on the surface is \(V = \frac{kQ}{R}\), so \(kQ = VR\). At \(r > R\), the electric field is \(E = \frac{kQ}{r^2} = \frac{VR}{r^2}\).

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