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 of the sphere, where \( r > R \), behaves as if the entire charge is concentrated at the center of the sphere. This is a consequence of the spherical symmetry of the charge distribution, and we can apply Coulomb's law to calculate the electric field. Step 1: Potential on the Surface of the Sphere.
The potential \( V \) at the surface of the sphere is related to the charge \( Q \) on the sphere and the radius \( R \) of the sphere by the formula: \[ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R}, \] where: \( Q \) is the total charge on the sphere, \( R \) is the radius of the sphere, \( \epsilon_0 \) is the permittivity of free space. Rearranging the equation for \( Q \), we get: \[ Q = 4\pi\epsilon_0 \cdot V \cdot R. \] Step 2: Electric Field Outside the Sphere.
Now, we need to calculate the electric field \( E \) at a distance \( r \) from the center of the sphere, where \( r > R \). According to Coulomb's law, the electric field at a distance \( r \) from a point charge is given by: \[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2}. \] 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: Simplifying the Expression.
Simplifying the equation: \[ E = \frac{V \cdot R^2}{r^2}. \] Thus, the magnitude of the electric field at a distance \( r \) from the center of the sphere is: \[ E = \frac{V \cdot R^2}{r^2}. \] Therefore, the correct answer is \( \mathbf{(2)} \).