The electric field between the plates of a parallel-plate capacitor is horizonta

Question

The electric field between the plates of a parallel-plate capacitor is horizontal, uniform, and has a magnitude E=1.46×104N/C. A small object of mass 0.0250 kg and charge -3.10 ?C is suspended by a thread between the plates, as shown in the sketch.(Figure 1) The tension in the thread is 0.249 N and the thread makes an angle of 10.5 ? with the vertical.

Part A

Suppose the magnitude of the electric field is adjusted to give a tension of 0.253 N in the thread. This will also change the angle the thread makes with the vertical. Find the new value of E. (N/C)

Part B

Find the new angle between the thread and the vertical.

Explanation / Answer

There are a vertical force (mg) and a horizontal force (qE)
mg =0.0250*9.8= 0.245N
qE =1.46×104*3.1*10-6 =0.04526N
T = sqrt[mg^2+(qE)^2] = sqrt[0.245^2+0.04526^2] = 0.249N
arctan(qE/mg) = arctan(0.04526/0.245) = 10.5 degrees

Notice that qE does change but mg DOES NOT
0.253^2 = (qE)^2 + mg^2 => 0.253^2 - 0.245^2 = (qE)^2 => qE = 0.063
E = 0.063/3.1e-6 = 20322.6 N/C
arctan(0.063/0.245) = 14.4 degrees

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