Four parallel metal plates P1, P2, P3, and P4, each with an area of 7.50 cm², are separated successively by a distance of d = 1.19 mm, as shown in the figure below. Plate P1 is connected to the negative terminal of a battery, and P2 is connected to the positive terminal. The battery maintains a potential difference of 12.0 V.
(a) If P3 is connected to the negative terminal, what is the capacitance of the three-plate system P1, P2, P3?
(b) What is the charge on P2?
(c) If P4 is now connected to the positive terminal, what is the capacitance of the four-plate system P1, P2, P3, P4?
(d) What is the charge on P4?
The correct answer and explanation is :
(a) Capacitance of the Three-Plate System P1, P2, and P3
The system consists of three parallel plates, P1, P2, and P3, where P1 is connected to the negative terminal, P2 to the positive terminal, and P3 is connected to the negative terminal. This is effectively a series combination of two capacitors.
For the first capacitor (between P1 and P2):
- The area ( A = 7.50 \, \text{cm}^2 = 7.50 \times 10^{-4} \, \text{m}^2 )
- The distance between the plates ( d = 1.19 \, \text{mm} = 1.19 \times 10^{-3} \, \text{m} )
- The permittivity of free space ( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} )
The capacitance between plates P1 and P2 is given by the formula:
[
C_1 = \frac{\varepsilon_0 A}{d}
]
[
C_1 = \frac{(8.85 \times 10^{-12} \, \text{F/m})(7.50 \times 10^{-4} \, \text{m}^2)}{1.19 \times 10^{-3} \, \text{m}} = 5.58 \times 10^{-12} \, \text{F}
]
For the second capacitor (between P2 and P3), the distance ( d ) is also ( 1.19 \, \text{mm} ), so the capacitance between P2 and P3 is the same:
[
C_2 = 5.58 \times 10^{-12} \, \text{F}
]
Since these two capacitors are in series, the total capacitance ( C_{\text{total}} ) is given by:
[
\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}
]
[
\frac{1}{C_{\text{total}}} = \frac{1}{5.58 \times 10^{-12}} + \frac{1}{5.58 \times 10^{-12}} = \frac{2}{5.58 \times 10^{-12}}
]
[
C_{\text{total}} = \frac{5.58 \times 10^{-12}}{2} = 2.79 \times 10^{-12} \, \text{F}
]
(b) Charge on P2
The charge on the plates is the same for both capacitors, as they are in series. Using the formula ( Q = C_{\text{total}} \times V ), where ( V = 12.0 \, \text{V} ), we get:
[
Q = (2.79 \times 10^{-12} \, \text{F}) \times (12.0 \, \text{V}) = 3.35 \times 10^{-11} \, \text{C}
]
Thus, the charge on P2 is ( 3.35 \times 10^{-11} \, \text{C} ).
(c) Capacitance of the Four-Plate System P1, P2, P3, and P4
Now, when P4 is connected to the positive terminal, we have two additional capacitors:
- The capacitance between P3 and P4 (same area and distance as before) is also ( 5.58 \times 10^{-12} \, \text{F} ).
The system now consists of three capacitors in series: ( C_1 ), ( C_2 ), and the capacitance between P3 and P4, which is again ( 5.58 \times 10^{-12} \, \text{F} ). The total capacitance ( C_{\text{total}} ) is:
[
\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
]
[
\frac{1}{C_{\text{total}}} = \frac{1}{5.58 \times 10^{-12}} + \frac{1}{5.58 \times 10^{-12}} + \frac{1}{5.58 \times 10^{-12}} = \frac{3}{5.58 \times 10^{-12}}
]
[
C_{\text{total}} = \frac{5.58 \times 10^{-12}}{3} = 1.86 \times 10^{-12} \, \text{F}
]
(d) Charge on P4
Since the system is now three capacitors in series, the charge on each plate is the same. Using the total capacitance and the applied voltage, we can find the charge on P4:
[
Q = C_{\text{total}} \times V = (1.86 \times 10^{-12} \, \text{F}) \times (12.0 \, \text{V}) = 2.23 \times 10^{-11} \, \text{C}
]
Thus, the charge on P4 is ( 2.23 \times 10^{-11} \, \text{C} ).
Summary of Results:
- (a) The capacitance of the three-plate system is ( 2.79 \times 10^{-12} \, \text{F} ).
- (b) The charge on P2 is ( 3.35 \times 10^{-11} \, \text{C} ).
- (c) The capacitance of the four-plate system is ( 1.86 \times 10^{-12} \, \text{F} ).
- (d) The charge on P4 is ( 2.23 \times 10^{-11} \, \text{C} ).
Here is the diagram showing the four parallel metal plates, P1, P2, P3, and P4, along with their connections to the battery terminals.