An electric charge \(10^{-6} \, \mu C\) is placed at the origin (0, 0) of an X-Y coordinate system. Two points P and Q are situated at \((\sqrt{3}, \sqrt{3}) \, \text{mm}\) and \((\sqrt{6}, 0) \, \text{mm}\) respectively. The potential difference between the points P and Q will be:
0 V
\( \sqrt{6} \) V
\(\sqrt{3} \) V
3 V
The potential difference between two points P and Q due to a point charge Q is given by:
\[ \Delta V = KQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]
where
\[ r_1 = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{6} \quad \text{and} \quad r_2 = \sqrt{(\sqrt{6})^2 + 0^2} = \sqrt{6} \]
Since \(r_1 = r_2\), the potential difference is zero:
\[ \Delta V = 0 \]
The electrostatic potential is also known as the electric field potential, electric potential, or potential drop is defined as “The amount of work that is done in order to move a unit charge from a reference point to a specific point inside the field without producing an acceleration.”
SI unit of electrostatic potential - volt
Other units - statvolt
Symbol of electrostatic potential - V or φ
Dimensional formula - ML2T3I-1
The electric potential energy of the system is given by the following formula:
U = 1/(4πεº) × [q1q2/d]
Where q1 and q2 are the two charges that are separated by the distance d.