Question 413: Electrostatics & Capacitance

Question

A capacitor of unknown capacity is connected across a battery of V volt. The charge stored in it is Q coulomb. When potential across the capacitor is reduced by $\mathrm{V}_1$ volt, the charge stored in it becomes $\mathrm{Q}_1$ coulomb. The potential V is

Accepted Answer

Step 1: Formula for charge stored
The charge stored in a capacitor is given by the formula: $$Q = C V$$ where $Q$ is the charge, $C$ is the capacitance, and $V$ is the voltage.
Step 2: Charge after voltage decreases
When the voltage is reduced by $V_1$, the new voltage becomes $V - V_1$. The new charge stored is: $$Q_1 = C (V - V_1)$$
Step 3: Isolate $C$ (capacitance) from the second equation
We can find $C$ from the second equation: $$C = \frac{Q_1}{V - V_1}$$
Step 4: Substitute $C$ back into the first equation
Now, substitute $C$ into the first equation: $$Q = \left(\frac{Q_1}{V - V_1}\right)V$$
Step 5: Rearranging to solve for $V$
Multiply both sides by $(V - V_1)$: $$Q(V - V_1) = Q_1 V$$
Expand and bring terms involving $V$ to one side: $$Q V - Q V_1 = Q_1 V$$
Now, get all $V$ terms together: $$Q V - Q_1 V = Q V_1$$
Factor $V$ on the left: $$V (Q - Q_1) = Q V_1$$
Now, solve for $V$: $$V = \frac{Q V_1}{Q - Q_1}$$

Answer

$\frac{\mathrm{QV}_1}{\mathrm{Q}-\mathrm{Q}_1}$
X

Answer

$\frac{Q_1 V_1}{Q+Q_1}$

Answer

$\frac{Q_1}{Q}$

Answer

$\frac{Q}{Q_1}$

Electrostatics & Capacitance

A capacitor of unknown capacity is connected across a battery of V volt. The charge stored in it is Q coulomb. When potential across the capacitor is reduced by \mathrm{V}_1 volt, the charge stored in it becomes \mathrm{Q}_1 coulomb. The potential V is

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