Question

The electric field intensity on the surface of a charged solid sphere of radius $r$ and volume charge density $\rho$ is given by ($\varepsilon_0$ = permittivity of free space)

• A. zero
• B. $\frac{\rho r}{3\varepsilon_0}$
• C. $\frac{1}{4\pi\varepsilon_0}\frac{\rho}{r}$
• D. $\frac{5\rho}{6\pi\varepsilon_0}$

Hint:

Dimensional analysis can verify this immediately! Electric field units are derived from $\frac{\text{Charge}}{\varepsilon_0 \times \text{Area}}$. Since charge density $\rho$ is $\frac{\text{Charge}}{\text{Volume}}$, multiplying $\rho$ by a length parameter $r$ leaves you with exactly the correct $\frac{\text{Charge}}{\text{Area}}$ dimensions in the numerator, confirming the form $\frac{\rho r}{\varepsilon_0}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the mathematical expression of the electrostatic field strength ($E$) evaluated exactly on the outer boundary surface of a uniformly charged solid sphere.
The sphere is defined by its physical radius $r$ and its uniform internal volume charge density $\rho$.

Step 2: Key Formula or Approach:
1. According to Gauss's Law, the electric field on or outside the surface of a spherically symmetric distribution behaves as though all the total charge $q$ is packed directly at its center:
$$E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$$ 2. The total charge $q$ contained within a solid sphere of radius $r$ can be expressed in terms of volume charge density $\rho$ multiplied by the sphere's geometric volume:
$$q = \rho \times V = \rho \times \left(\frac{4}{3}\pi r^3\right)$$

Step 3: Detailed Explanation:
Substitute the volume-charge identity for $q$ directly into the surface electric field expression:
$$E = \frac{1}{4\pi\varepsilon_0} \frac{\rho \left(\frac{4}{3}\pi r^3\right)}{r^2}$$ Let's group and cancel out matching mathematical constants and terms in the numerator and denominator:
The factor $4\pi$ cancels completely from both parts.
The radius terms simplify since $\frac{r^3}{r^2} = r$.
Rewriting the remaining factors yields:
$$E = \frac{\rho r}{3\varepsilon_0}$$

Step 4: Final Answer:
The electric field intensity on the surface is $\frac{\rho r}{3\varepsilon_0}$, which corresponds perfectly to option (B).