Question

The electric field intensity on the surface of a solid charged sphere of radius 'r' and volume charge density '$\rho$' is ($\epsilon_0 =$ permittivity of free space)

• A. $\frac{\rho r}{3\epsilon_{0}}$
• B. $\frac{\rho}{4\pi\epsilon_{0}r}$
• C. zero
• D. $\frac{5\rho r}{6\epsilon_{0}}$

Hint:

Physics Tip: Inside a uniformly charged solid sphere, the electric field increases linearly with the distance from the center. On the surface, it reaches the value derived here.

Answer 1

The Correct Option is A

Concept: Physics (Electrostatics) - Application of Gauss's Law.

Step 1: State Gauss's Law. The total electric flux ($\phi$) through a closed surface is equal to the enclosed charge ($q_{enc}$) divided by the permittivity of free space ($\epsilon_0$): $$\phi = E \cdot A = \frac{q_{enc}}{\epsilon_{0}} \text{}$$

Step 2: Determine the enclosed charge. The charge is distributed throughout the volume of the sphere. Enclosed charge is the product of volume charge density ($\rho$) and the volume of the sphere ($V$): \ $$q_{enc} = \rho \times \left( \frac{4}{3}\pi r^{3} \right) \text{}$$

Step 3: Apply Gauss's Law to the surface. The surface area ($A$) of the sphere of radius $r$ is $4\pi r^2$. Substituting the area and enclosed charge into the flux equation: $$E(4\pi r^{2}) = \frac{\rho \left( \frac{4}{3}\pi r^{3} \right)}{\epsilon_{0}} \text{}$$

Step 4: Solve for the electric field $E$. Canceling $4\pi r^2$ from both sides: $$E = \frac{\rho r}{3\epsilon_{0}} \text{}$$ $$ \therefore \text{The electric field intensity on the surface is } \frac{\rho r}{3\epsilon_{0}}. \text{} $$