(A) Find the electric field inthe region between the two plates. Since the field

Question

(A) Find the electric field inthe region between the two plates. Since the field is approximatelyuniform in the region between the plates, the work done moving asmall test charge q between is:

(B) Find the chargeQ. Recall in connection with Equation (24.9) that applyingGauss's law to a small cylinder with only one flat face of theGaussian surface outside the conductor leads to the conclusion thatthe electric field immediately outside a conductor with a surfacecharge (in C/m2) is:

(C) Find the capacitance ofthe parallel plates. The capacitance is:

Explanation / Answer

Given

Surface area of plates A = 0.79 m2

Distance between plates d = 0.80 cm = 0.8 x 10-2 m

Applied voltage V = 12 V

Solution

A)

Electric field E = V/d

E = 12 / 0.8 x 10-2

E = 1500 V/m

B)

Electric field E = /o

E = /o

V/d = /o

= V o /d

Q/A = V o /d

Q = AV o /d

Q = 0.79 x 12 x 8.85 x 10-12 / 0.8 x 10-2

Q = 1.0487 x 10-8 C

C)

Capacitance C = Q/V

C = 1.0487 x 10-8 / 12

C = 8.739 x 10-10 F

C = 0.0008739 x 10-6 F

We should enter the value as 0.0008739 or as 8.739E-4 since x 10-6 is already there

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