Question

At the eight corners of a cube of side $10\,{\text{cm}}$, equal charges each of value $10\,{\text{C}}$ are placed. The potential at the centre of the cube is
A. $83.14 \times {10^{11}}{\text{V}}$
B. $16.62 \times {10^{11}}{\text{V}}$
C. $1.66 \times {10^{11}}{\text{V}}$
D. $1662.7 \times {10^{11}}{\text{V}}$

Answer 1

First, recall the formula for potential at a point due to a charge. Here eight charges are there at eight corners of the cube, distance of centre from all eight corners of a cube is the same. So, find the potential at the centre due to one charge and then use this value to find the total potential at the centre.

Complete step by step answer:
Given, side of the cube, $d = 10\,{\text{cm}} = 0.1\,{\text{m}}$. Charge on each corner of the cube, $Q = 10\,{\text{C}}$. Potential at a point due to a charge is given by the formula,
$V = \dfrac{{kq}}{r}$ ………....(i)
where $q$ is the charge, $r$ is the distance between the charge and the point and $k = 9 \times {10^9}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 2}}}}$ is proportionality constant.

Potential due to all eight charges at the corner at the centre of the cube will be the same as the distance of the centre from all the corners of the cube are the same and the charges are also equal.We will now have to find the distance of the centre of the cube from one of the corners of the cube.Distance of the centre to one of the corners of the cube will be half of the diagonal of the cube passing through the centre of the cube.Diagonal of a cube passing through its centre is written as,
$D = \sqrt 3 a$ where $a$ is the side of the cube.
Here side of the cube is $d = 0.1\,{\text{m}}$ so, diagonal of the cube passing through its centre is,
$D = \sqrt 3 \times 0.1$
The distance of the centre of the cube to one of its corner will be,
$r = \dfrac{D}{2}$
Putting the value of $D$ we get,
$r = \dfrac{{\sqrt 3 \times 0.1}}{2}$
$\Rightarrow r = 0.05\sqrt 3 \,{\text{m}}$
Here the charge is $Q = 10\,{\text{C}}$ and $r = 0.05\sqrt 3 \,{\text{m}}$. Putting these values in equation (i), we get
$V = \dfrac{{k \times 10}}{{0.05\sqrt 3 \,}}$
Putting the value of $k = 9 \times {10^9}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 2}}}}$, we get
$V = \dfrac{{9 \times {{10}^9} \times 10}}{{0.05\sqrt 3 \,}}$
$\Rightarrow V = \dfrac{{9 \times {{10}^9} \times 10 \times 100}}{{5\sqrt 3 \,}}$
$\Rightarrow V = \dfrac{{9 \times {{10}^9} \times 2 \times 100}}{{\sqrt 3 \,}}$
$\Rightarrow V = \dfrac{{18 \times {{10}^{11}}}}{{\sqrt 3 \,}}$
$\Rightarrow V = 10.39\, \times {10^{11}}{\text{V}}$
This is the potential at the centre due to one charge at the corner. So, potential at the centre due to eight charges at each corner of the cube will be,
${V_{total}} = 8V$
Putting the value of $V$ we get,
${V_{total}} = 8 \times 10.39 \times {10^{11}}{\text{V}}$
$\therefore {V_{total}} = 83.138 \times {10^{11}}{\text{V}} \approx {\text{83}}{\text{.14}} \times {10^{11}}{\text{V}}$

Hence, the correct answer is option A.

Note: We were asked to find the potential at the centre, so we found out potential due to a single charge and used to find the total potential as the distance and charge were equal. But, if we are asked to find potential at a point other than the centre of the cube then we have to calculate the potential due to each charge separately and add them to find the total potential.