Two identical metal plates are given charges $q_1$ and $q_2$ ($q_2 < q_1$) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance 'C', the potential difference 'V' between the plates is \dots
Show Hint
- $\frac{q_1 - q_2}{C}$
- $\frac{q_1 + q_2}{C}$
- $\frac{q_1 - q_2}{2C}$
- $\frac{q_1 + q_2}{2C}$
The Correct Option is C
Solution and Explanation
When two plates are given charges $q_1$ and $q_2$, the charges redistribute. The charge on the inner surface of the plates (which determines the potential difference) is $\pm \frac{q_1 - q_2}{2}$.
Step 2: Formula Application:
Potential difference $V = \frac{Q}{C}$, where $Q$ is the charge on the inner face of the positive plate.
Step 3: Explanation:
The charge on the inner face is $Q_{inner} = \frac{q_1 - q_2}{2}$. Therefore, $V = \frac{(q_1 - q_2)/2}{C} = \frac{q_1 - q_2}{2C}$.
Step 4: Final Answer:
The potential difference is $\frac{q_1 - q_2}{2C}$.
Top Questions on Capacitors and Capacitance
- Energy stored in a capacitor is given by the equation \[ E = \frac{1}{2} C V^2 \] where: - \( C \) is the capacitance, - \( V \) is the voltage, - \( E \) is the energy stored. Given the values of \( C \), \( V \), and \( E \), determine the energy stored.}
- MHT CET - 2025
- Physics
- Capacitors and Capacitance
- Two capacitors $ C_1 = 4\mu F $ and $ C_2 = 6\mu F $ are connected in series across a 60 V battery. The potential difference across $ C_2 $ is:
- BITSAT - 2025
- Physics
- Capacitors and Capacitance
A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.
As soon as the circuit is completed by closing key S₁ (keeping S₂ open) charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference Vc (= q/C) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = VC). During this process of charging, the charge q on the capacitor changes with time t as
\(q = Q[1 - e^{-t/RC}]\)
The charging current can be obtained by differentiating it and using
\(\frac{d}{dx} (e^{mx}) = me^{mx}\)
Consider the case when R = 20 kΩ, C = 500 μF and V = 10 V.- CBSE Class XII - 2025
- Physics
- Capacitors and Capacitance
- A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current (\( I_c \)) and the displacement current (\( I_d \)) has the same value at all points of the circuit.
- CBSE Class XII - 2025
- Physics
- Capacitors and Capacitance
- Want to practice more? Try solving extra ecology questions todayView All Questions