Question

The work done in bringing charge \[q_2\] from infinity to Position 2, in the electric potential \[V_1\] produced by the fixed charge \[q_1\] at separation \[r_{12}\], is:

विकल्प

• \[W_2 = q_1 \times V_1 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1^2}{r_{12}}\]

• \[W_2 = q_2 \times V_1 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r_{12}}\]

• \[W_2 = q_2^2 \times V_1 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_2^2}{r_{12}}\]

• \[W_2 = q_2 \times V_1 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1}{r_{12}}\]

Answer 1

The work done in bringing \[q_2\] from infinity to Position 2 equals the charge multiplied by the potential at that location: \[W_2 = q_2 \times V_1 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r_{12}}\].