Problem 49 - Master Resource Book in JEE Main Physics

Question

A charge \(5 \mu \mathrm{C}\) is placed at a point. What is the work required to carry \(1 \mathrm{C}\) of charge once round it in circle of \(12 \mathrm{~cm}\) radius? (a) 100 (b) 0 (c) 1 (d) \(\infty\)

Step-by-Step Solution

Answer

The work required is 0.

Step 1: Understanding the Problem

We are asked to find the work required to move a charge around another fixed charge. The fixed charge is given as \(5 \mu \mathrm{C}\), and we need to move \(1 \mathrm{C}\) of charge around it in a circle of radius \(12 \mathrm{cm}\).

Step 2: Recall the Concept of Electric Potential

Work to move a charge from one point to another in an electric field depends on the potential difference between those points. Moving charge in a circle around a point charge means there is no change in potential energy, as the distance remains constant.

Step 3: Conclusion Based on Concepts

Since the potential at any point around a point charge at a constant radius (circle) is the same, no work is done, because there is no change in electric potential energy. Work is only done when there is a change in potential energy.

Key Concepts

Electric Potential

Electric potential is a measure of the potential energy per unit charge at a certain point in an electric field. It tells us how much potential energy a charge has due to its position in the field. The electric potential at a distance from a point charge is determined by the equation:\[V = \frac{kQ}{r}\]where:

• \( V \) is the electric potential,
• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) \),
• \( Q \) is the charge creating the electric field,
• \( r \) is the distance from the charge to the point where the potential is being measured.

Understanding electric potential is important because it gives us insight into how charges interact at a distance in an electric field. If a charge is moved in such a way that it remains at the same potential, like moving around in a circle around a point charge with no change in radius, there is no change in potential energy. Therefore, no work is done.

Work Done by Electric Field

Work done by an electric field is an important aspect of electrostatics. It is the energy required to move a charge within an electric field. The formula to calculate work done \( W \) when moving a charge \( q \) from one point to another is given by:\[W = q(V_b - V_a)\]where:

• \( V_b \) is the electric potential at the final point,
• \( V_a \) is the electric potential at the initial point.

The key idea here is that work is only done if there is a change in the electric potential (\( V_b - V_a \)). If you move a charge at a constant distance around a fixed charge, like in a circular path, you're not changing the electric potential. So in this scenario, the work done is zero. This is because the electric field does not perform any net work as the initial and final potential energies are the same.

Circular Motion in Electric Field

Moving a charge in a circular motion within an electric field can be very interesting as it involves steady distances from a central point charge. To illustrate this with an example, when a charge completes a circular path around another charge, it remains equidistant at all points along the path. This means the potential energy at every point on the circle is identical. As there is no change in distance from the central point charge, the electric potential energy also remains constant. Circular motion also introduces the concept of centripetal force in physics, but it’s crucial to recognize the difference when it comes to electric fields. In this context, no external work is required to maintain the motion unless there's a force perpendicular to the motion of the charge. In our specific case, because the charge is simply moved around a circle at a constant radius of 12 cm, there is no change in the electric potential difference, and hence, no work is done by or against the electric field. This principle illustrates why the answer to the textbook problem was zero.