Question

A uniformly charged conducting sphere of 4.4m diameter has a surface charge density of \[60\mu C{{m}^{-2}}\]. The charge on the sphere is then
A). \[7.3\times {{10}^{-3}}C\]
B). \[3.7\times {{10}^{-6}}C\]
C). \[7.3\times {{10}^{-6}}C\]
D). \[3.7\times {{10}^{-3}}C\]

Answer 1

Hint: To calculate the charge on the surface of the sphere, the formula to calculate surface charge density will be used. The surface charge density is given by the rate of change of charge with respect to surface area. From this we can calculate the total charge.

Formula used: \[dq=\sigma \times ds\]

Complete step-by-step solution -
For the entire sphere of surface area \[4\pi {{r}^{2}}\] the total charge would be:
\[Q=\sigma \times 4\pi {{r}^{2}}\]

Given that diameter of the sphere is=4.4m
The radius would be r=2.2m
Surface charge density,
\[\sigma =60\times {{10}^{-6}}C{{m}^{-2}}\]
Total charge on the surface of the sphere is the product of surface charge density and surface area of sphere and its given by:
\[Q=\sigma \times 4\pi {{r}^{2}}\]
\[Q=60\times {{10}^{-6}}\times 4\times \dfrac{22}{7}\times {{(2.2)}^{2}}\]
\[Q=3650.7\times {{10}^{-6}}C\]
\[Q=3.7\times {{10}^{-3}}C\]
The correct answer is option D. \[3.7\times {{10}^{-3}}C\]

Additional Information:
The electric field due to uniformly charged spherical shell can be calculated using Gauss’ Theorem according to which electric flux over a closed surface is equal to \[\dfrac{1}{{{\in }_{0}}}\] times the charge enclosed within the surface.
\[\phi =\dfrac{q}{{{\in }_{0}}}\]
And the electric field is:
\[E=\dfrac{q}{4\pi {{\in }_{0}}{{r}^{2}}}\].

Note: Students must remember that the electric field due to a uniformly charged thin spherical shell gives the same result as the electric field due to a point charge. Which implies that a uniformly charged shell behaves like a point charge.