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A parallel plate capacitor has a constant electric field between the plates of 1

ID: 2113607 • Letter: A

Question

A parallel plate capacitor has a constant electric field between the plates of 100,000 N/C. If the capacitor is comprised of plates 1 cm x 1 cm separated by a distance of 1 mm, what is the surface charge on each plate? Got Answer of:Given Electric field between the plates, E = 10,000 N/C Area of the plates, A = 1 cm2 Distance between the plates, d = 1mm Surface charge density on each plate, ? = Q / A ? = C V / A = (?0 A /d ) (E d) / A ? = ?0 E = (8.85 * 10 -12 C2 / N m 2) (10,000 N / C) ? = 8.85 *10^-8 C/m^2 Now: What is the electric field in the capacitor in the previous problem if a mylar sheet is placed between the plates?

Explanation / Answer

E=sigma/epsilon

sigma=epsilon*E

=8.85 * 10^-8

E1=E/k

=100000/3.2=31250


thus, new sigma=epsilon*E1

=2.765 * 10^-7 C/m^2