Electric Potential

The work done by an external agent in moving a charge between two points at different potentials is equal to the change in electric potential energy: (W_{ext} = q(V_f - V_i)), where (q) is the charge, (V_f) the final potential, and (V_i) the initial potential. This relation follows from the definition of electric potential as potential energy per unit charge. The external agent must do positive work to move a positive charge to a region of higher potential, overcoming the repulsion from existing field configuration. Here, (q = 3 \times 10^{-6}) C, (V_f = 500) V, (V_i = 200) V, so (\Delta V = 300) V. Work done: (W = 3 \times 10^{-6} \times 300 = 9 \times 10^{-4}) J (= 0.9) mJ. Option 0.6 mJ is incorrect because it uses (\Delta V = 200) V instead of 300 V. Option 1.5 mJ is incorrect because it uses (\Delta V = 500) V instead of the difference 300 V. Option (-0.9) mJ is incorrect because it confuses the work done by the electric force with the work done by the external agent; the electric force would do (-0.9) mJ, not the external agent. This directly applies the JEE Main concept of potential difference and work. Plausibility check: the positive sign confirms that the external agent does positive work to move a positive charge up the potential hill.