Question

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) \(\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]\) (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) \(\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]\) (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

Short answer

The work done in moving a charge \(+Q\) along the semicircle \(CRD\) is approximately \(\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]\), although none of the given answer choices precisely equal to zero. However, option (C) is the most similar to the correct value.

Step-by-step solution

1. Identify the given information

: We are given: - Charge \(+q\) at point A - Charge \(-q\) at point B - The distance between point A and B is \(2L\) - Point C is the midpoint between point A and point B - Charge \(+Q\) is moved along the semicircle \(CRD\)

2. Determine the electric field produced by the charges

: The electric field produced by a single charge is given by: \[E = \frac{k\cdot q}{r^2}\] where - \(E\) is the electric field - \(k\) is the electric constant, \(k = \frac{1}{4\pi\epsilon_0}\) - \(q\) is the charge - \(r\) is the distance between the charge and the point of interest

3. Consider the symmetry of the situation

: Since point C is the midpoint between point A and point B, the electric field from charge \(+q\) at point A and charge \(-q\) at point B are equal and opposite at point C. Hence, the electric fields at point C cancel each other and the net field is zero. As the charge \(+Q\) moves along the semicircle, the same cancellation occurs due to symmetry.

4. Calculate the work done

: Since the electric field at all points on the semicircle is zero due to the symmetry of the situation, the work done on charge \(+Q\) in moving along the semicircle \(CRD\) is also zero. Therefore, the work done is: \[W = 0\] Considering the given options, none of them is precisely equal to zero. However, among the options, (A) and (C) are positive, while (B) and (D) are negative. It can be interpreted that the options are missing a sign and thus, option (C) is the most similar to zero, so we can consider it to be the correct answer when compared to the other given choices.

Key concepts

Electric Field

Understanding the electric field is crucial in electrostatics. An electric field is a region around a charged particle where a force would be experienced by other charges. The strength and direction of the electric field are represented by vectors, pointing away from positive charges and toward negative charges. The formula to calculate the electric field \[ E \] is given by:

• \[ E = \frac{k \cdot q}{r^2} \]
• Where \( k \) is the electric constant \( = \frac{1}{4 \pi \epsilon_0} \)
• \( q \) is the charge creating the field
• \( r \) is the distance from the charge

By understanding this, we can determine the influence of electric charges at various points in space, helping to solve complex problems like the one in the step by step solution.

Work Done

In physics, work is defined as a force acting upon an object to cause displacement. In the context of electrostatics, work done involves moving a charge against an electric field. The work done on a charge \( +Q \) is the energy transferred to move it through an electric field. When the net electric field is zero, such as in the given scenario, the electric force does no work. Thus, the work done is:

• \[ W = 0 \]

This is because there is no net movement of electric potential energy to account for. If the electric field along the semicircle is zero due to symmetry, moving the charge does not require energy.

Electric Charge

Electric charge is a fundamental property of matter, associated with electromagnetic forces. Charges can be positive or negative, with like charges repelling and opposite charges attracting each other. In the given problem, we have charges \(+q\) and \(-q\), which create an electric field. Charge is measured in Coulombs, and the interactions between charges are described by Coulomb's law. Understanding charge interactions is vital in calculating fields, determining forces, and eventually finding work done in moving charges across electric fields.

Symmetry in Electric Fields

Symmetry plays a significant role in simplifying electric field problems. When two charges are equidistant from a point, like in the exercise where point C is midway between charges \(+q\) and \(-q\), symmetries simplify calculations. The electric fields due to each charge cancel each other out at points of symmetry, like point C and along the semicircle \(CRD\). This results in a net field of zero. Hence, any movement of a charge within this symmetric setup, as in the semicircle movement problem, incurs no net work. Utilizing symmetry helps identify when forces or fields might cancel, providing a clearer path to problem-solving in electrostatics.