5.10 Capacitor plates in two frames * A capacitor consists of two parallel recta

Question

5.10 Capacitor plates in two frames * A capacitor consists of two parallel rectangular plates with a ver tical separation of 2 cm. The east-west dimension of the plates is 20 cm, the north-south dimension is 10 cm. The capacitor has been charged by connecting it temporarily to a battery of 300 V What is excess electrons are on the negative plate? Now give the following quantities as they would be measured in a frame of reference that is moving eastward, relative to the laboratory in which the plates are at rest, with speed 0.6c: the three dimensions of the capacitor the number of excess electrons on the negative plate; the electric field strength between the plates. Answer the same questions for a the electric field strength between the plates? How many ame of reference that is moving upward with speed 0.6c.

Explanation / Answer

Electric field between the plates will be:

E = V/d = 300/0.02 = 1.5 x 10^4 V/m

Hence, E = 1.5 x 10^4 V/m

We know that capacitance is given by:

C = e0 A/d

A = 10 x 20 = 200 cm^2 = 0.02 m^2

C = 8.85 x 10^-12 x 0.02/0.02 = 8.85 x 10^-12 F

Q = CV = 8.85 x 10^-12 x 300 = 2.655 x 10^-9 C

we know that one electron has q = 1.6 x 10^-19.

So the number of electrons will be:

N = 2.655 x 10^-9/1.6 x 10^-19 = 1.66 x 10^10 electrons

Hence N = 1.66 x 10^10 electrons

Now in the moving reference frame:

we know from length contraction

L' = L sqrt (1 - v^2/c^2)

for v = 0.6c, the new dimensions of plates will be:

L'1 = 16 cm ; L'2 = 8 cm

A = 16 x 8 = 128 cm^2 = 0.0128 m^2

C = e0 A/d = 8.85 x 10^-12 x 0.0128/0.02 = 5.66 x 10^-12 F

Q = CV = 5.66 x 10^-12 x 300 = 1.698 x 10^-9 C

N = 1.698 x 10^-9/1.6 x 10^-19 = 1.06 x 10^10 electrons

Hence, N = 1.06 x 10^10 electrons

E = 1.5 x 10^4 N/C (remains same)

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