Question:medium

A charge \( -6 \, \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \, \mu C \) is moved from point C to point A along the circumference. Calculate the work done on the charge.
A charge −6µC is placed

Show Hint

The work done in moving a charge in an electric field depends on the potential difference between the initial and final points. If the potential is the same, no work is done.
Updated On: Feb 19, 2026
Show Solution

Solution and Explanation

1. Electric Potential from a Point Charge:

The electric potential \( V \) at a location generated by a point charge \( Q \) is calculated using the formula:

\[ V = \frac{kQ}{r} \]

Where:

  • \( k \) represents Coulomb's constant, with the value \(k = 9 \times 10^9 \, \text{N.m}^2/\text{C}^2\),
  • \( Q \) is the magnitude of the charge responsible for the potential,
  • \( r \) is the distance from the charge to the point of potential calculation.

2. Work Required to Move a Charge:

The work \( W \) expended to transfer a charge \( q \) between two points within an electric field is determined by:

\[ W = q \Delta V \]

In this equation:

  • \( q \) denotes the charge being relocated, and
  • \( \Delta V \) signifies the difference in electric potential between the starting and ending locations.

 

3. Electric Potential at Points A and C:

A charge of \( -6 \, \mu C \) is situated at the center (point B) of the semicircle. Consequently, the electric potential at any point on the semicircle, including points A and C, will be identical. This is because all points on the semicircle are equidistant from the central charge, as they lie on the radius.

Therefore, the electric potential at points A and C is equivalent due to their equal distances from the central charge \( -6 \, \mu C \).

4. Work Done Calculation:

Given that the electric potential at both points A and C is identical, the potential difference \( \Delta V \) between them is zero. Consequently, the work done in moving the charge \( +5 \, \mu C \) from point C to point A is calculated as follows:

\[ W = q \Delta V = 5 \, \mu C \times 0 = 0 \]

5. Final Determination:

  • The net work performed on the charge \( +5 \, \mu C \) during its movement from point C to point A along the semicircular path is zero.
Was this answer helpful?
2