Physics - Capacitor Question with Solution | TestHub

The capacitance $C$ of a capacitor is given by the formula:
$C=\frac{Q}{V}$

From the given graph of voltage ($V$) versus charge ($Q$), the slope of the line is:
$\text{Slope}=\frac{V}{Q}=\frac{1}{C}$

From the graph, it is clear that the slope of line B is greater than the slope of line A.
${\text{Slope}}_B>{\text{Slope}}_A$
$\Rightarrow \frac{1}{C_B}>\frac{1}{C_A}$
$\Rightarrow C_A>C_B$

Alternatively, for a given constant charge $Q$, the graph shows that $V_B>V_A$.
Since $C\propto \frac{1}{V}$ for a constant $Q$, a smaller voltage corresponds to a higher capacitance.
Therefore, $C_A>C_B$, which means the capacity of A is higher than B.

Answer: capacity of $A$ is higher than $B$.