The drawing shows a parallel plate capacitor that is moving with a speed of 32 m

Question

The drawing shows a parallel plate capacitor that is moving with a speed of 32 m/s through a 3.6-T magnetic field. The velocity V is perpendicular to the magnetic filed. The electric field within the capacitor has a value of 170 N/C, and each plate has an area of 7.5 times 10^-4 m^2. Calculate the magnitude of the magnetic force exerted on the positive plate. At a certain location, the horizontal component of the earth's magnetic field is 2.5 times 10^-5 T, due north. A proton moves eastward with just the right speed for the magnetic force on it to balance its weigh. Compute the speed of the proton. [mass of protron = 1.67 times 10^-27 kg, charge = 1.602 times 10^-19 C, g = 9.8 m/s^2]

Explanation / Answer

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

Q = (8.85x10-12)(7.5x10-4)*(170) = 1.128 x10-12 C

F = q v B sin = (1.128 x10-12 C)*(32m/s)*(3.6 T) = 1.29*10-10 N

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