(C) Find the capacitance of the parallel plates. Please answer C how is this unc

Question

(C) Find the capacitance of the parallel plates.

Please answer C

how is this unclear? Please answer THE LAST BOX, thanks.

Analyze (A) Find the electric field in the region between the two plates Since the field is approximately uniform in the region between the two plates, the work done moving a small test charge q between them is which is also the change in potential energy So equate these two expressions to find 8.0V=1000 (1) E =-- 0.0080 m V/m. (B) Find the charge Q Applying Gauss's law to a small cylinder with only one flat face of the Gaussian surface outside the conductor leads to the conclusion that the electric field immediately outside a conductor with a surface charge (in C/m2) is So the charge on each plate has magnitude Use Equation (1) in Equation (2) ( ( A = (8.85 x 10-12)08.0V--(0.99 m,2 8.7615e-9 = (C) Find the capacitance of the parallel plates. The capacitance is which can be evaluated from the previously calculated results as x 10-6 F

Explanation / Answer

Using Q= 8.7615e-9 and V=8V; C= Q/V= e0*A/d= 8.85e-12*0.99/0.0080= 1.095*10^-9= 0.00195 * 10^-6F

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