With the information given, write the equation of the capacitance in terms of ep

Question

With the information given, write the equation of the capacitance in terms of epsilon_0 A, d. Solve for the numerical value of C in pF. With the information given express the potential difference across the capacitor in terms of Q and C. Solve for the memorial value of DeltaV in V If the charges on the plates are increased to 2Q, what is the value of the capacitance C in pF? If A is them to 2A d is decreased to 4 what is the value of the capacitance C in pF? What is the new value of difference Delta Vm V with the Q Two parallel pates are charged with +Q and -Q respectively as shown in the figure, where Q = 25n C. The area of each plate is A = 0.029 m^2. The distance between them is d = 9.2 cm. The plates are in the air. With the information given write the equation of the capacitance in terms of z0 A d

Explanation / Answer

Here ,

A = 0.029 m^2

Q = 25 nC

d = 9.2 cm = 0.092 m

part A)

as the capacitance is given as

C = Area * epsilon/seperation

C = A*epsilon/d

the capacitance is A*E0/d

part B)

putting in values

C = 0.029 * 8.854 *10^-12/.092

C = 2.791 *10^-12 F = 2.791 pF

the capacitance is 2.791 pF

C)

as the charge in the capacitor is given as

Q = C * V

V = Q/C

the potential is Q/C

D)

V = Q/C

V = 25 *10^-9/(2.791 *10^-12 )

V = 8957 V

the potential difference is 8957 V

e)

as the capacitance is indeopendent on charge

capacitance will be same as before

C = 2.791 *10^-12 F = 2.791 pF

the capacitance is 2.791 pF

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