Question 464: Electrostats & Capacitance

Question

The graph shows the variation of voltage (v) across the plates of two parallel plate capacitos $A$ and $B$ versus increase of charge $Q$ stored in them. Then

Accepted Answer

In the graph, $\mathrm{V}_{\mathrm{A}}$ and $\mathrm{V}_{\mathrm{B}}$ are the voltages across parallel plate capacitors A and B corresponding to charge Q on each of the capacitors.
$\mathrm{V}_{\mathrm{A}}=\mathrm{Q} / \mathrm{C}_{\mathrm{A}}$ and $\mathrm{V}_{\mathrm{B}}=\mathrm{Q} / \mathrm{C}_{\mathrm{B}}$
or $\frac{V_B}{V_A}=\frac{\frac{Q}{C_B}}{\frac{Q}{C_A}}=\frac{C_A}{C_B}$
Since, $V_B>V_A, \frac{V_B}{V_A}=\frac{C_A}{C_B}>1$
$	herefore \quad C_A>C_B$, i.e. capacitor $A$ has the higher capacitance.

Answer

capacity of both capacitors is same.

Answer

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

Answer

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

Answer

capacity of both is zero.

Electrostats & Capacitance

The graph shows the variation of voltage (v) across the plates of two parallel plate capacitos A and B versus increase of charge Q stored in them. Then

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