Goal Calculate fundamental physical properties of a parallel-plate capacitor. Pr

Question

Goal Calculate fundamental physical properties of a parallel-plate capacitor. Problem A parallel-plate capacitor has an area A =2.50 x 10-4 m2 and a plate separation d = 1.20 x 10-3 m. Find its capacitance. How much charge is on the positive plate if the capacitor is connected to a 3.00 V battery? Calculate the charge density on the positive plate, assuming the density is uniform. Calculate the magnitude of the electric field between the plates. Strategy Parts (a) and (b) can be solved by substituting into the basic equations for capacitance. In part (c), use the fact that the voltage difference equals the electric field times the distance. Find the capacitance. Substitute into Equation 16.9. C = F Find the charge on the positive plate after the capacitor is connected to a 3.00 V battery. Substitute into Equation 16.9. Calculate the charge density on the positive plate. Charge density is charge divided by area Calculate the magnitude of the electric field between the plates. Apply Equation 15.13. Remarks The answer to part (d) could also have been obtained from the electric potential, which is AV = Ed for a parallel-plate capacitor.

Explanation / Answer

A. AE/d=2.5*10^-4*8.85*10^-12/1.2*10^-3=1.84375pF b. Q=CV=1.84375810^-12*3=5.53*10^-12C C. sigma=5.53*10^-12/2.50*10^-4 =2.212*10^-8C/m^2 d. E=2.212*10^-8/8.85*10^-12 =2499.4N/C

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