A condenser of 2 μ F capacitance is charged steadily from 0 to 5 C. Which of the following graph represents correctly the variation of potential difference (V) across it's plates with respect to the charge (Q) on the condenser?

Solution

Physics sir · 3 steps

C stays fixed while charge goes in, so V = Q/C is a straight line through the origin. Find where it ends.

1. Shape of graph. The capacitance depends only on the plates and the medium between them, not on how much charge is stored. So C stays at 2 μF throughout, and V is directly proportional to Q. The graph is a straight line from the origin.

   V = Q/C = (5 C)/(2 × 10⁻⁶ F) = 2.5 × 10⁶ V

2. End point. When Q = 5 C, the voltage is 2.5 × 10⁶ V. The slope of the line is 1/C = 5 × 10⁵ V C⁻¹.
3. So. The correct graph is a straight line from (0, 0) to (5 C, 2.5 × 10⁶ V). That is option (a).

The idea to keep. For a capacitor, V = Q/C with C fixed by geometry, so V against Q is always a straight line with slope 1/C. Only the end point needs any arithmetic.

Answer from the NTA final answer key. Solution by PrepWiser.