Question

The surface density on the copper sphere is $\sigma$. The electric field strength on the surface of the sphere is-
(A). $\sigma$
(B). $\dfrac{\sigma }{2}$
(C). $\dfrac{\sigma }{2{{\varepsilon }_{0}}}$
(D). $\dfrac{\sigma }{{{\varepsilon }_{0}}}$

Answer 1

Assume the total charge on the sphere is concentrated at the center; Its Gaussian’s surface will be a sphere. Equate electric flux by definition and by gauss law. Solve the equation to calculate the electric field. Substitute the charge in terms of surface charge density of copper.
Formulas used:
$\phi =EA$
$\phi =\dfrac{Q}{{{\varepsilon }_{0}}}$
$\sigma =\dfrac{Q}{4\pi {{r}^{2}}}$

Complete step-by-step solution
All charge on the copper sphere is assumed to be concentrated in the center of the sphere. Let us assume a Gaussian surface around the charge of the sphere then the flux passing through the sphere is given by Gauss law as-
$\phi =EA$ - (1)
Here, $\phi$is the total flux passing through the surface
$E$ is the electric field due to the charge
$A$ is the area of the cross-section.
Also, we know that,
$\phi =\dfrac{Q}{{{\varepsilon }_{0}}}$ - (2)
$Q$ is the total charge
${{\varepsilon }_{0}}$ is the permeability of free space
From eq (1) and eq (2), we get,
$\dfrac{Q}{{{\varepsilon }_{0}}}=EA$
$A=4\pi {{r}^{2}}$ ($r$ is the radius of sphere)
$\Rightarrow \dfrac{Q}{{{\varepsilon }_{0}}}=E(4\pi {{r}^{2}})$
$\therefore E=\dfrac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$
Therefore, the electric field due to the sphere is $\dfrac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$
The surface density is the charge per unit surface area, therefore, surface density of copper is-

& \sigma =\dfrac{Q}{4\pi {{r}^{2}}} \\\ & \Rightarrow Q=\sigma 4\pi {{r}^{2}} \\\ \end{aligned}$$ Substituting $$Q$$ in electric field we get, $$\begin{aligned} & E=\dfrac{\sigma 4\pi {{r}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}} \\\ & \therefore E=\dfrac{\sigma }{{{\varepsilon }_{0}}} \\\ \end{aligned}$$ **The electric field of a copper sphere with density $$\sigma $$ is $$\dfrac{\sigma }{{{\varepsilon }_{0}}}$$. Therefore, the correct option is (D).** **Note:** The number of electric lines of forces passing through a surface is called electric flux. The electric flux is given by- $$\phi =\overrightarrow{E}\cdot \overrightarrow{S}$$ ($$\overrightarrow{E}$$ is the electric field vector and $$\overrightarrow{S}$$ is the area vector). The Gauss law states that the total flux passing through a closed area is equal to the total charge enclosed in the area divided by the permittivity of the surrounding material.