The two plates of a capacitor hold +2500 \; \mu C and -2500 \; \mu C of charge, respectively, when the potential difference is 960 V. What is the capacitance?

This is Giancoli Answers with Mr. Dychko. The charge collected on each plate of a capacitor is equal to the capacitance times the potential difference across the plates and divide both sides by V and solve for C, capacitance is the charge 2500 times ten to the minus six coulombs divided by the voltage of 950 volts until the capacitance is about 2.6 micro farads.

COMMENTS

By huggingallthetrees on Mon, 2/24/2020 - 8:18 AM

per the given verbiage, the potential difference is 960 V, not 950 V.
Same calculation, though!
Thanks Mr. Dychko!

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The two plates of a capacitor hold +2500 μC+2500 \; \mu \textrm{C}+2500μC and −2500 μC-2500 \; \mu \textrm{C}−2500μC of charge, respectively, when the potential difference is 960 V. What is the capacitance?

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The two plates of a capacitor hold $+2500 \; \mu \textrm{C}$ and $-2500 \; \mu \textrm{C}$ of charge, respectively, when the potential difference is 960 V. What is the capacitance?