Problem 118 - A Complete Resource Book in Physics for JEE Main

Question

Work done by electric field, when particle moves from \(C_{A}\) to \(C_{B}\) is (A) \(-1.2 \mathrm{~J}\) (B) \(1.2 \mathrm{~J}\) (C) \(-3.6 \mathrm{~J}\) (D) \(3.6 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done by the electric field is \(-1.2 \; \mathrm{J}\).

Step 1: Identifying known variables

From the given problem, we identify the known quantities. The charge of the particle \(q\) is \(6 \times 10^{-3} \; \mathrm{C}\) and the potential difference (or voltage) between points \(C_{A}\) and \(C_{ B}\) is \(200 \; \mathrm{V}\).

Step 2: Applying the formula for work done

The formula for the work done \(W\) by an electric field is \(W = qV\), where \(q\) is the charge and \(V\) is the voltage or potential difference.

Step 3: Calculating the work done

Substitute the identified quantities into the equation: \(W = (6 \times 10^{-3} \; \mathrm{C})(200 \; \mathrm{V}) = 1.2 \; \mathrm{J}\).

Step 4: Determining the direction of the work done

Work done by the electric field is considered negative because the electric field does work to move a particle against its field, therefore the work done by the electric field is \(-1.2 \; \mathrm{J}\).