Question

Four charges each of 1 \[\mu \]C are placed at the corners of a square of side \[2\sqrt{2}m\], the potential at the point of intersection of the diagonals is?
(A) \[18000V\]
(B) 1800 V
(C) \[18\sqrt{2}\times {{10}^{3}}V\]
(D) none of the above

Answer 1

Hint: Four charges all of them have the same magnitude and same polarity. They are placed at the corners of the square whose side is \[2\sqrt{2}m\]. We need to find the electric potential at the intersection point of the diagonals. We know diagonals of a square bisect each other at right angles.

Complete step by step answer:
Magnitude of each charge, q= \[1\mu C={{10}^{-6}}C\]
Given the side of square, a=\[2\sqrt{2}m\]
To calculate the diagonal of a square d, using the formula,
\[\begin{align}
&\Rightarrow d=a\sqrt{2} \\
&\Rightarrow d=2\sqrt{2}\times \sqrt{2} \\
&\Rightarrow d =4m \\
\end{align}\]
Since they bisect each other so the distance between the point of intersection of the two diagonals from each vertex is 2 m.
We know electric potential is a scalar quantity given by the formula,
\[\Rightarrow V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{r}\]and here \[\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\] is a constant.
Since all the charges are of the same magnitude and at same distance so net potential at the centre will be
\[\therefore V=4(\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{r})=\dfrac{4\times 9\times {{10}^{9}}\times {{10}^{-6}}}{2}=18000V\].
So, the potential at the centre comes out to be 18000 V.

Hence,the correct option is (A).

Note:Here we have used the property of a square that it has all the four sides equal and the diagonals are also equal and they bisect each other and that at right angles. Then we have to calculate the potential due to each charge and add to get net potential. Electric potential energy is the energy that is needed to move a charge against an electric field. The electrical potential is a scalar quantity.