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 . A small object of mass 0.0250 and charge -3.10 is suspended by a thread between the plates, as shown in the sketch.(Figure 1) The tension in the thread is 0.249 and the thread makes an angle of 10.5 with the vertical.
Suppose the magnitude of the electric field is adjusted to give a tension of 0.254 in the thread. This will also change the angle the thread makes with the vertical. Find the new value of .

Explanation / Answer

I assume the system is not accelerating. i found the angle of the string relative to the horizontal first and got 76 degrees. Calculate the y component of the tension force with (.35N)sin76=.3396035N. Because the system is not accelerating, all forces are equal. The y component of the tension force =gravity force because the two forces cancel each other. gravity force=(mass)(gravity's acceleration). Set gravity and tension equal to each other solve for mass. I got 34.653 g. Calculate the x component of the tension force: (.35N)cos76=.08467266N. The x component equals the force of the electric field acting on the charge: Force of electric field on charge=(charge)(Electric field magnitude). .08467266N=-2.15*10^-6(Electric field magnitude)

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