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

Hint: 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-
\[\begin{align}
  & \sigma =\dfrac{Q}{4\pi {{r}^{2}}} \\
 & \Rightarrow Q=\sigma 4\pi {{r}^{2}} \\
\end{align}\]
Substituting \[Q\] in electric field we get,
\[\begin{align}
  & E=\dfrac{\sigma 4\pi {{r}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}} \\
 & \therefore E=\dfrac{\sigma }{{{\varepsilon }_{0}}} \\
\end{align}\]
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.