The drawing shows a parallel plate capacitor that is moving with a speed of 30 m/s through a 3.1 T magnetic field. The velocity vector v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 170 N/C, and each plate has an area of 7.5 10-4 m2. What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

asked by Dee

11 years ago

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1 answer

E = Q/εoA, => Q = εo AE

Q = 220(8.85x10-12)(8x10-4) = 1.5576 x10^-12 C

F = q v B sin θ = 1.56(26)3.1E-12 = 125.7E -12 N

answered by Anonymous

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Question

The drawing shows a parallel plate capacitor that is moving with a speed of 30 m/s through a 3.1 T magnetic field. The velocity vector v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 170 N/C, and each plate has an area of 7.5 10-4 m2. What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

1 answer

E = Q/εoA, => Q = εo AE

Q = 220(8.85x10-12)(8x10-4) = 1.5576 x10^-12 C

F = q v B sin θ = 1.56(26)3.1E-12 = 125.7E -12 N

Anonymous · Answer

E = Q/εoA, => Q = εo AE Q = 220(8.85x10-12)(8x10-4) = 1.5576 x10^-12 C F = q v B sin θ = 1.56(26)3.1E-12 = 125.7E -12 N