Read the following paragraphs and answer the questions that follow. A parallel
plate capacitor consists of two conducting plates kept generally parallel to each other at
a distance. When the capacitor is charged, the charge resides on the inner surfaces of
the plates and an electric field is set up between them. Thus, electrostatic energy is
stored in the capacitor.
The figure shows three large square metallic plates, each of side ‘L’, held parallel and
equidistant from each other. The space between P1 and P2, and P2 and P3 is completely
filled with mica sheets of dielectric constant ‘K’.
The plate P2 is connected to point A and the other plates P1 and P3 are connected to
point B. Point A is maintained at a positive potential with respect to point B and the
potential difference between A and B is V .
Question: 1
The capacitance of the system between A and B will be:
Show Hint
When multiple capacitors are connected in parallel, add their capacitances directly: \( C_{\text{eq}} = C_1 + C_2 + \cdots \)
- The system is composed of two capacitors in parallel: one between $P_1$ and $P_2$, and the other between $P_2$ and $P_3$.
- Each has plate area \( A = L^2 \), separation \( d \), and dielectric constant \( K \).
- Capacitance of one: \( C_1 = \frac{K\varepsilon_0 L^2}{d} \)
- Similarly, \( C_2 = \frac{K\varepsilon_0 L^2}{d} \)
- Since they are in parallel (same potential across both):
\[
C_{\text{total}} = C_1 + C_2 = \frac{K\varepsilon_0 L^2}{d} + \frac{K\varepsilon_0 L^2}{d} = \frac{2K\varepsilon_0 L^2}{d}
\]
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Question: 2
The charge on plate \( P_1 \) is:
Show Hint
Charge on a capacitor is given by \( Q = CV \). For systems with dielectrics, use \( C = \frac{K\varepsilon_0 A}{d} \).
- As discussed earlier, the system behaves like two capacitors in parallel.
- The potential difference across each capacitor is \( V \).
- Capacitance between \( P_1 \) and \( P_2 \): \( C_1 = \frac{\varepsilon_0 K L^2}{d} \)
- Therefore, the charge on \( P_1 \):
\[
Q = C_1 \cdot V = \frac{\varepsilon_0 K L^2}{d} \cdot V = \frac{\varepsilon_0 V K L^2}{d}
\]
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Question: 3
The electric field in the region between \( P_1 \) and \( P_2 \) is:
Show Hint
Electric field between two plates with a potential difference \( V \) and separation \( d \) is given by \( E = \frac{V}{d} \).
- The total potential difference between points A and B is \( V \).
- Since the dielectric is placed and the system consists of two capacitors in series with equal spacing \( d \), the potential drop across each capacitor is \( \frac{V}{2} \).
- But in this setup, plates \( P_1 \) and \( P_3 \) are both at the same potential \( B \), while \( P_2 \) is at potential \( A \).
- So the potential difference between \( P_1 \) and \( P_2 \) is \( V \), and the distance between them is \( d \).
- Thus, the electric field \( E = \frac{V}{d} \)
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Question: 4
The separation between the plates of same area (\( L^2 \)) of a parallel plate air capacitor having capacitance equal to that of this system, will be:
Show Hint
When comparing dielectric and air capacitors with equal capacitance, remember: increasing the dielectric constant \( K \) allows a larger capacitance at the same plate separation—or the same capacitance with a larger separation.
- The equivalent capacitance of the given setup is:
\[ C = \frac{2 \varepsilon_0 K L^2}{d} \]
- Let the air capacitor have the same plate area \( A = L^2 \) and plate separation \( x \), and it should have the same capacitance.
- Capacitance of air capacitor:
\[ C' = \frac{\varepsilon_0 A}{x} = \frac{\varepsilon_0 L^2}{x} \]
- Equating the two:
\[
\frac{\varepsilon_0 L^2}{x} = \frac{2 \varepsilon_0 K L^2}{d} \Rightarrow \frac{1}{x} = \frac{2K}{d} \Rightarrow x = \frac{d}{2K}
\]
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Question: 5
If the source of potential difference applied between A and B is removed, and then A and B are connected by a conducting wire, the net charge on the system will be:
Show Hint
When a capacitor is disconnected from the battery and both terminals are joined by a wire, the system is isolated and neutral — thus, total net charge is always zero.
- Initially, when the source of potential \( V \) is connected between A and B, charges are induced on the plates and an electric field is established.
- Once the source is removed, the system becomes isolated.
- Connecting A and B with a conducting wire makes them equipotential.
- Since the system is now closed and isolated with no external source, no net charge can exist in the entire system.
- Charge gets redistributed internally but the total net charge remains zero.
