A voltmeter connected to a fully charged parallel plate 6.00nF capacitor reads 1

Question

A voltmeter connected to a fully charged parallel plate 6.00nF capacitor reads 18.0V. The capacitor is not connected to a battery and the distance between the plates is 2.00mm. A dielectric is placed between the plates of the capacitor and the voltmeter then reads 6.00V. Without the dielectric what are (a) the charge on the plates, and (b) the electric field strength between the plates. With the dielectric between the plates what are (c) the dielectric constant, (d) the electirc field strength, (e) the capacitance, and (f) the charge on the plates.

Explanation / Answer

C = capacitance = 6 nF = 6 x 10-9 F

V = voltage without dielectric = 18 V

d = distance between plates = 2 mm = 2 x 10-3 m

V' = voltage with dielectric = 6 V

a) without dielectric

using the formula

Q = CV

Q = (6 x 10-9 ) (18)

Q = 0.108 x 10-6 C

B)

electric field between the plates is given as ::

E = V/d = 18 / ( 2 x 10-3 )

E = 9000 V/m

C)

with dielectric constant, let the dielectric constant be ''K''

Ck = new capacitance= K C

using the formula

Q = CkV'

charge remains same

0.108 x 10-6 = (6 x 10-9 ) k (6)

k = 3

d)

E' = V'/d

E' = 6 / (2 x 10-3 )

E' = 3000 V/m

e)

Ck = k C

Ck = 3 x 6 nF

Ck = 18 nF

f)

charge remains same

so Q = 0.108 x 10-6C

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