Question:

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:

Updated On: Nov 21, 2025
  • 0 V

  • \( \sqrt{6} \) V

  • \(\sqrt{3} \) V

  • 3 V

Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Approach Solution - 1

To find the potential difference between points P and Q due to a charge placed at the origin, we use the formula for electric potential due to a point charge:

\(V = \frac{kQ}{r}\) 

where \(V\) is the electric potential, \(k\approx8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\) is the Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance from the charge.

  1. Calculate the distance from the charge at the origin to point P \((\sqrt{3}, \sqrt{3}) \, \text{mm}\): \(r_P = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{6} \, \text{mm} = \sqrt{6} \times 10^{-3} \, \text{m}\)
  2. Determine the distance from the charge to point Q \((\sqrt{6}, 0) \, \text{mm}\): \(r_Q = \sqrt{(\sqrt{6})^2 + 0^2} = \sqrt{6} \, \text{mm} = \sqrt{6} \times 10^{-3} \, \text{m}\)
  3. Since both points P and Q are equidistant from the origin, the potentials at P and Q are the same: \(V_P = \frac{k \cdot 10^{-6} \cdot 10^{-6}}{\sqrt{6} \times 10^{-3}} = V_Q\)
  4. Thus, the potential difference between the two points P and Q is: \(\Delta V = V_P - V_Q = 0 \, \text{V}\)

The correct answer is 0 V. The potential at two equidistant points from the same charge is equal, leading to a potential difference of zero.

Was this answer helpful?
2
4
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

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:

\[ \Delta V = 0 \]

Was this answer helpful?
0
0

Top Questions on electrostatic potential and capacitance

View More Questions

Concepts Used:

Electrostatic Potential

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:

SI unit of electrostatic potential - volt

Other units - statvolt

Symbol of electrostatic potential - V or φ

Dimensional formula - ML2T3I-1

Electric Potential Formula:

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.