The drawing shows a parallel plate capacitor that is moving with a speed of 27 m

Question

The drawing shows a parallel plate capacitor that is moving with a speed of 27 m/s through a 4.3-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 9.9

The drawing shows a parallel plate capacitor that is moving with a speed of 27 m/s through a 4.3-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 9.9 times 10-4 m2. What is the magnitude of the magnetic force exerted on the positive plate of the capacitor?

Explanation / Answer

E = Q/?oA, => Q = ?o AE

Q = 130(8.85*10^-12)(9.9*10^-4) = 1.1389*10^-12 C

F = q v B sin ? = 1.1389*(27)*4.3*10^-12 = 132.22*10^-12 N

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