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Answer :
Final answer:
The parallel-plate capacitor's capacity is calculated by dividing the charge by the voltage. Upon applying correct conversions, it's found to be 4.00 x 10^-7 F in units of farads and 400 nF in units of nanofarads.
Explanation:
The capacitance of a capacitor, denoted by the symbol C, is defined by the relationship Q = CV where Q represents the charge stored on the capacitor, V is the voltage applied across the capacitor, and C is the capacitance. This equation can be rearranged to solve for capacitance, giving C = Q/V. In this case, the charge Q is 2.00 x 10^2 μC and the voltage V is 5.00 x 10^2 V. Note that we need to convert the charge from μC to C before performing the calculation. So the charge in coulombs is 2.00 x 10^2 x 10^-6 C = 2.00 x 10^-4 C. Using these values in our equation, we find C = 2.00 x 10^-4 C / 5.00 x 10^2 V = 4.00 x 10^-7 F or 0.4 μF. Since 1 μF equals 10^3 nF, the capacitance also equals 400 nF. Hence the parallel-plate capacitor's capacity is 4.00 x 10^-7 F in units of farads and 400 nF in units of nanofarads.
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