3. A parallel plate capacitor has an electric field of 160 N/C and each plate ha

Question

3. A parallel plate capacitor has an electric field of 160 N/C and each plate has an area of 7.4 cm2 . The capacitor is moving through a magnetic field, of strength 3.5 T, at a velocity of 30 m/s.    The electric and magnetic fields are pointed in the same direction.

(a) What is the electrical force between the plates?

(b) If the velocity of the capacitor is directed perpendicular to the magnetic field, how great is the magnetic force between the plates?

(c) If the velocity of the capacitor is directed so that the NORMAL to the plates is is moving parallel to the magnetic field, how great is the magnetic force between the plates?

Explanation / Answer

here,

Electric field, E = 160 N/C
area, a = 7.4 cm^2 = 0.00074 m^2
magnatic field, B = 3.5 T
velocity, V = 30 m/s

Charge on plates, Q = eo*a*E = 8.85*10-12 * 0.00074 * 160 = 87.079 C

Part A:
Electric force, Fe = q*E = 87.079 * 160 = 13932.64 N

Part b:
magnatic Force, Fb = q*V*B*Sin90
Fb = 87.079 * 30 * 3.5
Fb = 9143.295 N

Part C:
Since SinA = 0 degrees as velocity and field is parallel,

magnatic Force = q*v*B*Sin0 = 0 N

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