Choose the correct answer: A parallel-plate capacitor has circular plates of radius 8 cm and plate separation I mm. What will be the charge on the plates if a potential difference of 100 V is applied across its plates?

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1.78 $\times$ 10-8C Explanation: The capacitance of a parallel plate capacitor is given as follows: C = (ε_0A)/d Here, ε0 is the permittivity of free space, A is the plate area, and d is the separation between the plates. For circular plates as shown in the above figure, the area of the plates can be calculated as follows: A = piR^2 Therefore the capacitance can be expressed as: C = (ε_0 piR^2)/d .....(1) Substitute 8 cm for R, 1 mm for d, and 8.85 xx 10^-12 CV^-1 m^-1 for ε0 in equation (1) to find the value of C. C = `((8.85 xx 10^-12 CV^-1 m^-1)*(3.14)*(8 xx 10^-2 m^2)^2)/(1 xx 10^-3 m) C = 1.78 xx 10^-10 CV^-1` The charge on the capacitor is given as follows: Q = CV .....(2) Substitute 100V for V and 1.78 xx 10^-10 CV^-1 for C in equation (2) to find the value of Q. Q = `(1.78 xx 10^-10 CV^-1) * (100V) Q = 1.78 xx 10^-8 C`

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