Question

The voltage across a parallel plate capacitor of capacitance $127\mu \text{F}$ is increased from $3.887\text{V}$ to $4.887\text{V}$. The extra charge that flows through the battery is

• A. $63.5\mu \text{C}$
• B. $254\mu \text{C}$
• C. $508\mu \text{C}$
• D. $127\mu \text{C}$

Answer 1

The correct option is D $127\mu \text{C}$

Given:
Potential difference, $V=\delta V=4.887-3.887=1\text{V}$

Now, the equation, $C=\frac{Q}{V}$ states that capacitance is the charge stored when the voltage across the capacitor increases by $1\text{V}$.

$\Rightarrow 127\mu \text{F}=\frac{Q}{1\text{V}}\Rightarrow Q=127\mu \text{C}$

So, in this case the extra charge that gets stored is also the extra charge that flows through the battery is $127\mu \text{C}$.

Why this question ? Caution : If you applied or planning to apply $Q=CV$, to find the initial & final charge and then take their difference to arrive at charge flown, then you are bound to waste a lot of time & this process is inefficient.