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(a)
Whether the capacitors are connected in series or parallel.
(a)
Answer to Problem 46P
The two capacitors are connected in the parallel connection.
Explanation of Solution
Introduction:
The combination of the capacitors consider as the parallel combination if the potential difference across all the capacitors is same. If the charge through the each capacitor remains same then combination is called the series combination of capacitors.
One end of all the capacitors is connected by the positive terminal of the battery and other end of two capacitors connected to negative terminal of the battery, so the potential difference across two plates remain same. As the potential difference across the two capacitors plate is same, so both capacitors connected in the parallel connection.
Conclusion:
Thus, the two capacitors are connected in the parallel connection.
(b)
The equivalent capacitance of the capacitors.
(b)
Answer to Problem 46P
The equivalent capacitance as a function area of plates and distance between the plates is
εA(1d1+1d2) .
Explanation of Solution
Given info:
The distance between the middle plate and upper plate is, d1 .
The distance between the middle plate and lower plate is, d2 .
The area of each plate is, A .
Formula Used:
The expression to calculate the capacitance of first capacitor is given as,
C1=εAd1
Here,
ε is the permittivity of space between the plates.
The expression to calculate the capacitance of second capacitor is given as,
C2=εAd2
The expression to calculate the charge of first capacitor is,
Q1=C1V
Here,
V is the potential difference across each plates.
The expression to calculate the charge of second capacitor is,
Q2=C2V
The expression to calculate the total charge of the equivalent connection is,
Q=CV
The capacitors are connected in parallel, so the potential across each capacitor is same and total charge of both capacitors is equal to sum of charge of each capacitor.
The expression to calculate the total charge of the both capacitors in parallel is given by,
Q=Q1+Q2 ..........(1)
Calculation:
Substitute the given values in the expression (1).
CV=C1V+C2VC=C1+C2C=εAd1+εAd2C=εA(1d1+1d2)
Conclusion:
Thus, the equivalent capacitance as a function area of plates and distance between the plates is εA(1d1+1d2) .
Chapter 19 Solutions
Physics: Principles with Applications
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