Two 2.90 cm x 2.90cm plates that form a parallel-plate capacitor are charged to

Question

Two 2.90 cm x 2.90cm plates that form a parallel-plate capacitor are charged to +/- 0.708 nC.

A) What is electric field strength inside the capacitor if the spacing between the plates is 1.50 mm?

B) What is the potential difference across the capacitor if the spacing between the plates is 1.50 mm?

C) What is the electric field strength inside the capacitor if the spacing between the plates is 3.00 mm?

D) What is the potential difference across the capacitor if the spacing between the plates is 3.00 mm?

Part A You may want to review (Pae $78-683 What is the electric field strength inside the capecitor if the spacing betwreen the plates is 1.50 mm? Part 8

Explanation / Answer

a) Use the Equation E= (Q/A)/Esub0
Area = .029m x .029m = 8.41×10^-4
Q = 0.708×10^-9
Q/A = 0.708×10^-9/8.41×10^-4= 8.41 x10^-7
Q/A/Esub0=8.41 x10^-7/8.85×10^-12=95124 V/m

b)

V = E*d

= 95124*1.50*10^-3

= 142.686 volts

c) Distance does not chang E

d)

V = E*d

= 95124*3.0*10^-3

= 285.372 volts

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