A condenser of $$2 \mu F$$ capacitance is charged steadily from $$0$$ to $$5 \text{ C}$$. Which of the following graph represents correctly the variation of potential difference $$V$$ across its plates with respect to the charge $$Q$$ on the condenser?

We need to determine the correct graph representing the variation of potential difference ($$V$$) across a condenser's plates with respect to the charge ($$Q$$) on it.

1. Relate Potential Difference ($$V$$) and Charge ($$Q$$)

The fundamental relationship between the charge stored in a capacitor, its capacitance ($$C$$), and the voltage across its plates is given by:

$$Q = C \times V$$

Rearranging the equation to solve for $$V$$ (since $$V$$ is plotted on the vertical $$y$$-axis and $$Q$$ is plotted on the horizontal $$x$$-axis):

$$V = \left(\frac{1}{C}\right) \cdot Q$$

This matches the equation of a straight line passing through the origin ($$y = mx$$), where:

• The slope of the line is $$m = \frac{1}{C}$$.
• Since capacitance $$C$$ is a positive constant, the slope is a constant positive value, making the graph a straight, linearly rising line starting from $$(0,0)$$.

2. Calculate the Coordinates at Maximum Charge

From the problem 

• Capacitance ($$C$$) = $$2\ \mu\text{F} = 2 \times 10^{-6}\text{ F}$$
• Maximum charge ($$Q$$) = $$5\text{ C}$$

Substitute these values into our equation to find the corresponding maximum voltage:

$$V = \frac{5}{2 \times 10^{-6}} = 2.5 \times 10^6\text{ V}$$

The vertical axis on the graph is scaled by a factor of $$10^6\text{ V}$$. Therefore, the value shown on the $$y$$-axis at $$Q = 5\text{ C}$$ must be $$2.5$$.

Conclusion

The correct graph is a straight line starting from zero and reaching a peak coordinate of $$(5, 2.5)$$. This matches Option A perfectly.