35. (I) The two plates of a capacitor hold +2500 µC and -2500 μC of charge, respectively, when the potential dif- ference is 960 V. What is the capacitance?
35. (I) The two plates of a capacitor hold +2500 µC and -2500 μC of charge, respectively, when the potential dif- ference is 960 V. What is the capacitance?
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please type your solution so that it is easy for to read I have bad eyesight please and thank you
![### Problem 35: Capacitor Charge and Capacitance
The two plates of a capacitor hold charges of +2500 μC and -2500 μC, respectively, when the potential difference between them is 960 V. The question asks to determine the capacitance of the capacitor.
### Solution:
To find the capacitance \( C \) of the capacitor, use the formula:
\[ C = \frac{Q}{V} \]
Where:
- \( Q \) is the charge on one of the plates (since the charge on the plates is equal and opposite, we can use the magnitude of the charge),
- \( V \) is the potential difference between the plates.
Given data:
- \( |Q| = 2500 \, \mu C = 2500 \times 10^{-6} \, C \)
- \( V = 960 \, V \)
Substitute the known values into the capacitance formula:
\[ C = \frac{2500 \times 10^{-6} \, C}{960 \, V} \]
Calculate the capacitance:
\[ C = \frac{2500 \times 10^{-6}}{960} \, F \]
\[ C \approx 2.60 \times 10^{-6} \, F \]
So, the capacitance \( C \) is approximately 2.60 μF (microfarads).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F100ca580-01b3-4065-a77f-0a16a29f1100%2F3ad247b9-396f-47d5-8c8c-382ae23f5fc6%2F09f3nnf_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 35: Capacitor Charge and Capacitance
The two plates of a capacitor hold charges of +2500 μC and -2500 μC, respectively, when the potential difference between them is 960 V. The question asks to determine the capacitance of the capacitor.
### Solution:
To find the capacitance \( C \) of the capacitor, use the formula:
\[ C = \frac{Q}{V} \]
Where:
- \( Q \) is the charge on one of the plates (since the charge on the plates is equal and opposite, we can use the magnitude of the charge),
- \( V \) is the potential difference between the plates.
Given data:
- \( |Q| = 2500 \, \mu C = 2500 \times 10^{-6} \, C \)
- \( V = 960 \, V \)
Substitute the known values into the capacitance formula:
\[ C = \frac{2500 \times 10^{-6} \, C}{960 \, V} \]
Calculate the capacitance:
\[ C = \frac{2500 \times 10^{-6}}{960} \, F \]
\[ C \approx 2.60 \times 10^{-6} \, F \]
So, the capacitance \( C \) is approximately 2.60 μF (microfarads).
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