Question

A metallic sphere of radius \( R \) is charged to a potential \( V \). The magnitude of the electric field at a distance \( r \, (r > R) \) from the center of the sphere is:

• A. \( \frac{V}{r^2} \)
• B. \( \frac{VR^2}{r^2} \)
• C. \( \frac{V}{R^2} \)
• D. \( \frac{V}{r} \)

Hint:

For a metallic sphere, the electric field outside the sphere behaves as if the entire charge is concentrated at the center. To calculate the electric field, use the formula \( E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} \), where \( Q \) is the charge and \( r \) is the distance from the center of the sphere.

Answer 1

The Correct Option is B

For a charged metallic sphere, the electric field at a distance \( r \) from the center, where \( r>R \), is equivalent to that of a point charge located at the sphere's center. This is due to the spherical symmetry of the charge distribution, allowing the application of Coulomb's law.Step 1: Surface Potential Calculation.
The potential \( V \) on the sphere's surface is determined by its charge \( Q \) and radius \( R \) via the formula:\[V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R},\]where: \( Q \) denotes the total charge, \( R \) is the sphere's radius, \( \epsilon_0 \) represents the permittivity of free space.Solving for \( Q \):\[Q = 4\pi\epsilon_0 \cdot V \cdot R.\]Step 2: External Electric Field Determination.
The electric field \( E \) at a distance \( r \) from the center (\( r>R \)) is calculated using Coulomb's law for a point charge:\[E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2}.\]Substituting the expression for \( Q \) from Step 1:\[E = \frac{1}{4\pi\epsilon_0} \cdot \frac{4\pi\epsilon_0 \cdot V \cdot R}{r^2}.\]Step 3: Expression Simplification.
Simplifying the derived equation yields:\[E = \frac{V \cdot R^2}{r^2}.\]Thus, the magnitude of the electric field at distance \( r \) is:\[E = \frac{V \cdot R^2}{r^2}.\]Therefore, the correct answer is \( \mathbf{(2)} \).