Two parallel plates, each having area A = 3250 cm^2 are connected to the termina

Question

Two parallel plates, each having area A = 3250 cm^2 are connected to the terminals of a battery of voltage V_b = 6 V as shown. The plates are separated by a distance d = 0.45 cm. What is Q, the charge on the top plate? What is U, the energy stored in the this capacitor? The battery is now disconnected from the plates and the separation of the plates is doubted (= 0.9 cm). What is the energy stored in this new capacitor? What is E, the magnitude of the electric field in the region between the plates? Compare V, the magnitude of the new potential difference across the plates, to V_b, the voltage of the battery. V V_b Two uncharged parallel plates are now connected to the initial pair of plates as shown. How will the electric field, E, and potential difference across the plates, V, change, if at all? Both E and V will remain the same E will decrease and V will increase E will increase and V will decrease Both E and V will decrease Both E and V will increase

Explanation / Answer

(1)

charge Q = C*V

C = capacitance = eo*A/d = 6.4*10^-10 F

Q = C*V = 3.835*10^-9 C

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(2)

energy stored = (1/2)*Q^2/C = (1/2)*(3.835*10^-9)^2/(6.4*10^-10)

U = 1.15*10^-8 J

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(3)

C1 = eo*A/(2d) = C/2

U1 = (1/2)*Q^2/C1 = 2*U = 1.3*10^-8 J

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(4)

E = Q/(A*eo)

E = 3.385*10^-9/(3250*10^-4*8.85*10^-12)

E = 1176.8 N/C

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5)

Vb = E*d

V = E*2d = 2*Vb

v > vb <<<<<----------answer

==============

(6)

Both E and V will decrease

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