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# The figure represents a square carrying charges +q, +q, -q, -q at its corners as shown. Then the potential will be zero at points:A) A, B, C, P, and QB) A, B, and CC) A, P, C, and QD) P, B, and Q

Last updated date: 01st Mar 2024
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Hint: To solve the above question you should be familiar with the electric potential energy. Suppose an electric charge \$q\$ is brought into an electric field \$E\$ . Since the electric field is energy conservative therefore the work done will be stored as the potential energy. This potential energy is called electric potential energy. And we express this energy as \$U = \dfrac{{kq}}{r}\$ , where \$k\$ is coulomb constant. Use this formula to find the points of zero potential.

Complete step by step solution:
First let us calculate potential at point A. the above figure is a charge-system of four charges. To calculate the net potential at A we will add the potential at A due to \$ + q, + q, - q\$ and \$ - q\$ with the help of the above formula. Let us label each charge in the figure. But remember that every charge is of an equal amount. The sign may differ.

If you look at the figure you will see that the distance between A and \${q_2}\$ , and distance between A and \${q_4}\$ are equally let this distance is \${r_2}\$. Also, A is in the middle of \${q_1}\$ and \${q_3}\$ , let this distance is \${r_1}\$ .
Therefore the net potential at A will be
\$\therefore {V_A} = \dfrac{{k{q_1}}}{{{r_1}}} + \dfrac{{k{q_3}}}{{{r_1}}} + \dfrac{{k{q_2}}}{{{r_2}}} + \dfrac{{k{q_4}}}{{{r_2}}}\$
Since every charge has equal magnitude, therefore substitute the value \$q\$ with their sign.
\[\therefore {V_A} = \dfrac{{kq}}{{{r_1}}} - \dfrac{{kq}}{{{r_1}}} + \dfrac{{kq}}{{{r_2}}} - \dfrac{{kq}}{{{r_2}}}\]
\$ \Rightarrow {V_A} = 0\$ .
Therefore net potential at A is zero.
Similarly, point C has the same conditions as point A, there also the potential will be zero.
Point B is in the middle of the square where each corner is at an equal distance. Let this distance is \$r\$ , then
\$\therefore {V_B} = \dfrac{{k{q_1}}}{r} + \dfrac{{k{q_2}}}{r} + \dfrac{{k{q_3}}}{r} + \dfrac{{k{q_4}}}{r}\$
Since every charge has equal magnitude, therefore substitute the value \$q\$ with their sign.
\[\therefore {V_B} = \dfrac{{kq}}{r} + \dfrac{{kq}}{r} - \dfrac{{kq}}{r} - \dfrac{{kq}}{r}\]
\$ \Rightarrow {V_B} = 0\$ .
Therefore the potential at B is zero.
Now let us discuss the potential at P. We see that the distance between a negative charge and point P is more than that between a positive charge and point P. Therefore the net potential at point P will not be zero. And if you see clearly the same condition holds for point Q. Therefore at point Q also the potential will not be zero.

Therefore the potential will be zero at A, B, and C. Hence option B is the correct option.

Note: Potential energy is a scalar quantity. Therefore we add this quantity with scalar addition rules. The potential energy is inversely proportional to the distance between two charges therefore if the distance increases potential decreases and if the distance decreases potential increases. This is the reason why the potential at P and Q is not zero.