Question

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10⁻⁹ C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Answer 1

Potential at point P = \(\frac q{4\pi \in_0d_1} = V_1\)

Potential at point Q = \(\frac q{4\pi \in_0d_2} = V_2\)

\(W = q_1[V_2 - V_1]\)

\(= q_1 \left[\frac q{4\pi \in_0d_2 }-\frac q{4\pi \in_0d_1} \right]\)

\(= \frac {qq_1}{4\pi \in _0} \left[\frac 1{d_2 }- \frac 1{d_1}\right]\)

\(= 9 \times 10^9 \times 8 \times 10^{-3} \times (-2 \times 10^{-9}) \left[\frac 1{0.04} - \frac 1{0.03}\right]\)

\(= -144\times 10^{-3} \times \left[\frac{-25}3\right]\)

\(= 1.27 J\)

\(\therefore \) Work done = 1.27J

Answer 2

Charge located at the origin, q = 8 mC= 8 × 10⁻³ C Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = − 2 × 10⁻⁹ C All the points are represented in the given figure.

Therefore, work done during the process is 1.27 J.