The graph shows the variation of potential difference between the plates of two capacitors A and B with the increase of charge stored by them. From the graph we can conclude:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Capacitance of A is greater than B
Relation between charge and potential difference:
$Q = CV \quad \Rightarrow \quad V = \frac{Q}{C}$
In $V$–$Q$ graph, slope = $\frac{V}{Q} = \frac{1}{C}$
From the graph, slope of B is greater than slope of A.
$\Rightarrow \frac{1}{C_B} \gt \frac{1}{C_A} \quad \Rightarrow \quad C_A \gt C_B$
Therefore, capacitance of A is greater than capacitance of B.