A parallel plate capacitor of capacitance 600\(\mu F\) has a potential difference of 200V between its plates. Calculate the charge on either plate of the ca...

Assessment: WAEC SSCE - Physics - 2005 (Objective) Subject: Physics

Question 1 Report

A parallel plate capacitor of capacitance 600\(\mu F\) has a potential difference of 200V between its plates. Calculate the charge on either plate of the capacitor
Answer Details
The charge on either plate of a capacitor is given by the formula: Q = C × V Where Q is the charge, C is the capacitance, and V is the potential difference between the plates of the capacitor. In this problem, the capacitance of the parallel plate capacitor is given as 600\(\mu F\) and the potential difference between the plates is given as 200V. Therefore, using the above formula, we can calculate the charge on either plate as follows: Q = C × V = 600 × 10^-6 F × 200 V = 0.12 C So the charge on either plate of the capacitor is 0.12 C, which is equal to 1.2 coulombs (since 1 coulomb is equal to 1 C). Therefore, the answer is: 1.2C.

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