Answered

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What is the capacitance of a capacitor that has 7.9 x 107 C of charge and a potential
difference of 12.0 V?

Sagot :

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Answer:

6.58×10⁶ F

Explanation:

Applying,

Q = CV..................... Equation 1

Where Q = charge of the capacitor, C = Capacitance of the capacitor, V = Volatage

Make C the subject of the equation

C = Q/V................. Equation 2

From the question,

Given: Q = 7.9×10⁷ C, V = 12.0 V

Susbtitute these values into equation 2

C = 7.9×10⁷/12

C = 6.58×10⁶ F

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