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.
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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.
The electric potential \( V \) at a point due to a point charge \( Q \) is given by the formula:
\[ V = \frac{kQ}{r} \]
Where:
\( k \) is Coulomb's constant, \(k = 9 \times 10^9 \, \text{N.m}^2/\text{C}^2\),
\( Q \) is the charge creating the potential,
\( r \) is the distance from the charge to the point where the potential is being calculated.
2. Work Done in Moving a Charge:
The work \( W \) done in moving a charge \( q \) from one point to another in an electric field is given by:
\[ W = q \Delta V \]
Where:
\( q \) is the charge being moved, and
\( \Delta V \) is the change in electric potential between the initial and final positions.
3. Electric Potential at Points A and C:
The charge \( -6 \, \mu C \) is at the center B of the semicircle. The potential at any point on the semicircle due to this central charge will be the same, as the distance from the center (radius) is constant for both points A and C.
Thus, the electric potential at points A and C is the same because both points are equidistant from the central charge \( -6 \, \mu C \).
4. Work Done:
Since the electric potential at both points A and C is the same, the potential difference \( \Delta V \) is zero. Therefore, the work done in moving the charge \( +5 \, \mu C \) from point C to point A is:
\[ W = q \Delta V = 5 \, \mu C \times 0 = 0 \]
5. Conclusion:
The work done on the charge \( +5 \, \mu C \) as it moves from point C to point A along the semicircular path is zero.
