Question

The drawing shows a parallel plate capacitor that is moving with a speed of 35 m/s through a 4.2-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 230 N/C, and each plate has an area of 8.8 × 10 ^-4 m ². What is the magnitude of the magnetic force exerted on the positive plate of the capacitor?

Chapter 21, Problem 09 The drawing shows a parallel plate capacitor that is moving with a speed of 35 m/s through a 4.2-T magnetic field. The velocity v is to the magnetic field. The electric field within the capacitor has a value of 230 N/C, and each plate has an area of 8.8 x 10-4 m2. What is the magnitude of the magnetic force exerted on the positive plate of the capacitor? Number Units the tolerance is +/-2% Click if you would like to show work for this question Open Show Work

Answer 1

Electric field inside a parallel plate capacitor can be given as,

$E = \frac{Q}{A\epsilon }$

Where E is the electric field; Q is the charge on each of the plates of the capacitor and epsilon is the permittivity of the medium.

Now from this relation, charge on each plate can be calculated as,

$Q = A\epsilon E$

$Q = 8.8\times 10^{-4} \times 8.854\times 10^{-12}\times 230 = 1.792\times 10^{-12}C$

Then magnetic force on the positive plate will be,

$F = BQv \sin 90^{0}$

$F =4.2\times 1.792\times 10^{-12}\times 35= 2.634\times 10^{-10}N$