Question

The electric field intensity on the surface of a solid charged sphere of radius \( r \) and volume charge density \( \sigma \) is \( (\varepsilon_0 = \text{permittivity of free space}) \)

• A. zero
• B. \( \frac{\sigma}{\varepsilon_0} \)
• C. \( \frac{1}{4 \pi \varepsilon_0} \frac{\sigma}{r^2} \)
• D. \( \frac{1}{3 \varepsilon_0} \)

Hint:

Gauss's law is a powerful tool for calculating electric fields for symmetric charge distributions, such as spherical, cylindrical, or planar symmetries.

Answer 1

The Correct Option is B

Step 1: Electric field from Gauss's Law.
According to Gauss's Law, the electric field \( E \) on the surface of a spherical object with a uniform charge distribution is given by:
\[ E = \frac{Q}{4 \pi \varepsilon_0 r^2}, \]
where \( Q \) is the total charge enclosed, \( \varepsilon_0 \) is the permittivity of free space, and \( r \) is the radius of the sphere.

Step 2: Total charge on the sphere.
The total charge \( Q \) on a sphere of radius \( r \) with a uniform surface charge density \( \sigma \) is related to the surface area \( A = 4 \pi r^2 \) of the sphere by:
\[ Q = \sigma \cdot A = \sigma \cdot 4 \pi r^2. \]

Step 3: Substituting the value of \( Q \) in Gauss's law.
Substituting the expression for \( Q \) into Gauss's law:
\[ E = \frac{\sigma \cdot 4 \pi r^2}{4 \pi \varepsilon_0 r^2} = \frac{\sigma}{\varepsilon_0}. \]

Step 4: Conclusion.
Thus, the electric field intensity on the surface of the solid charged sphere is:
\[ \boxed{\frac{\sigma}{\varepsilon_0}}. \]