Concept explainers
A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 250 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy of the capacitor.
(a)
The charges on plates before and after immersing in water.
Answer to Problem 66P
The charges on plates before and after immersing in water is
Explanation of Solution
Write the equation for capacitance of the system.
Here,
Write the equation for charge stored in the capacitor by using (I).
Since the voltage supply is disconnected before immersing in water, charge on the plates is same before and after immersion.
Conclusion:
Substitute,
Thus, the charges on plates before and after immersing in water is
(b)
The capacitance and potential difference after immersion.
Answer to Problem 66P
The capacitance and potential difference after immersion is
Explanation of Solution
Write the equation for capacitance of the system after immersion.
Here,
Write the expression for the potential difference after immersion.
Here,
Conclusion:
Substitute,
Substitute,
Thus, the capacitance and potential difference after immersion is
(c)
The change in energy of the capacitor.
Answer to Problem 66P
The change in energy of the capacitor is
Explanation of Solution
Write the expression for the initial capacitance.
Here,
Write the expression for the initial energy by using (V).
Write the expression for the final energy by using (III) and (IV).
Write the expression for the change in energy by using (VI) and (VII).
Conclusion:
Substitute,
Thus, the change in energy of the capacitor is
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Chapter 20 Solutions
Principles of Physics: A Calculus-Based Text
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