Two parallel plates, each having area A = 2126 cm 2 are connected to the termina

Question

Two parallel plates, each having area A = 2126 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.42 cm.

The battery is now disconnected from the plates and the separation of the plates is doubled ( = 0.84 cm).

What is E, the magnitude of the electric field in the region between the plates?

Two parallel plates, each having area A = 2126 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.42 cm. The battery is now disconnected from the plates and the separation of the plates is doubled ( = 0.84 cm). What is E, the magnitude of the electric field in the region between the plates?

Explanation / Answer

capaictance=epsilon*area/d=448.2 pF

so charge=C*V=2.689 nC

when separation is doubled,charge remains constant as voltage source is disconnected.

and capacitance is halved.

so

as voltage=charge/capacitance

voltage becomes doubled

so field=voltage/distance=2*6/(2*0.42*0.01)=1428.57 N/C

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