- #1

Jessehk

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## Homework Statement

A ping-pong ball of mass [tex]3.0 \times 10^{-15}[/tex] is hanging from a light thread 1.0 m long, between two vertical parallel plates 10 cm apart, as shown. When the potential difference across the plates is 420 V, the ball comes to equilibrium 1.0 cm to one side of its original position.

(See attached images for diagrams)

a) What is the electric field intensity between the plates?

b) What is the tension in the thread?

c) What is the magnitude of the electric force deflecting the ball?

d) What is the charge on the ball?

## Homework Equations

[tex]\xi = \frac{v}{d}[/tex]

[tex]F_e = \xi \cdot q[/tex]

## The Attempt at a Solution

First, I calculated the field strength between the two parallel plates.

[tex]\xi = \frac{420}{0.1} = 4200 N/C[/tex]

Then, I drew the attached free body diagram, and from it, I got the following:

[tex]cos \theta = \frac{0.01}{1}[/tex]

[tex]\theta \approx 89^{\circ{}}[/tex]

Then [tex]sin(89) = \frac{mg}{T}[/tex]

so [tex]T = 2.94 \times 10^{-3} N[/tex]

(which is the right answer).

However, I'm lost when I try to do part C. Any answer that I get is nonsensical.

(The answer in the book is [tex]2.94 \times 10^{-5} N[/tex])

I suspect that my FBD is not correct. I know that as soon as I know what Fe is, I can solve for the charge, q.

Did I also make a mistake in part A?

Any help would be greatly appreciated. If possible, It would be fantastic if somebody could point out the error in my thinking rather than just providing the solution. Thanks again.

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