The drawing shows a parallel plate capacitor that is moving with a speed of 25 m

Question

The drawing shows a parallel plate capacitor that is moving with a speed of 25  m/s through a 3.4-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 130  N/C, and each plate has an area of 9.7  

The drawing shows a parallel plate capacitor that is moving with a speed of 25 m/s through a 3.4-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 130 N/C, and each plate has an area of 9.7 times 10-4 m2. What is the magnitude of the magnetic force exerted on the positive plate of the capacitor?

Explanation / Answer

Q=CV =(eoA/d)(Ed) =eoAE

Q=8.85*10^-12*9.7*10^-4*130

Q=1.116*10^-12 C

F=QVB =1.116*10^-12*25*3.4

F=9.486*10^-11 N

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