Question

A cube of side 5 cm has a charge of \[6\,\mu C\]. The surface charge density in \[\mu C/{m^2}\] is:
A. 400
B. 200
C. 800
D. 100

Answer 1

Hint: Assume the length of the side of the cube is a and determine the total surface area of the cube. The surface charge density is the ratio of charge on the object to the total surface area of the object. Express the surface charge density in terms of \[\mu C/{m^2}\].

Formula used:
Surface charge density, \[\sigma = \dfrac{q}{A}\] ,
where, q is the charge and A is the surface area.

Complete step by step answer:
We have given that the length of the side of cube is \[a = 5\,cm = 0.05\,{\text{m}}\] and the charge on the cube is \[q = 6\,\mu C\].Let us calculate the surface area of the cube. The area of the side of the cube of length \[a = 0.05\,{\text{m}}\] is,
\[A = {a^2} = {\left( {0.05} \right)^2}\]
\[ \Rightarrow A = 2.5 \times {10^{ - 3}}\,{{\text{m}}^2}\]
But the cube has six sides. Therefore, the total surface area of the cube will be \[6A\].
\[A = 6 \times \left( {2.5 \times {{10}^{ - 3}}\,{{\text{m}}^2}} \right)\]
\[ \Rightarrow A = 0.015\,{{\text{m}}^2}\]
We know that the surface charge density of the object is equal charge per unit surface area.
\[\sigma = \dfrac{q}{A}\]
Here, q is the charge and A is the surface area.
Substituting \[q = 6\,\mu C\] and \[A = 0.015\,{{\text{m}}^2}\] in the above equation, we get,
\[\sigma = \dfrac{6}{{0.015}}\]
\[ \therefore \sigma = 400\,\mu C/{m^2}\]
Therefore, the surface charge density of the cube is \[400\,\mu C/{m^2}\].

So, the correct answer is option A.

Note:We were already asked to determine the surface charge density in \[\mu C/{m^2}\] so do not convert it into \[C/{m^2}\]. You have to calculate the total surface area of the cube and not the surface area of only one side of the cube. Since the cube has six sides, you have to multiply the area of one side of the cube by 6.