-1
$\begingroup$

Diagram

Suppose you have two equal point charges of magnitude $q$ and mass $m$ suspended by strings of length $L$ on opposite sides of a pendulum, both making an angle of theta with the $y$-axis. The question asks to demonstrate that the charge of the particles are given by $$q=2L\sin{\theta}\sqrt{\frac{mg\tan{\theta}}{k_e}}$$

But what I got was

(1) $F_g=F_e$

(2) $mg\sin{\theta}=\frac{q^2k_e}{r^2}$

(3) $mg\sin{\theta}=\frac{q^2k_e}{(2L\sin{\theta})^2}$

(4) $q=2L\sin{\theta}\sqrt{\frac{mg\sin{\theta}}{k_e}}$

Which obviously has the incorrect trig function inside the radical, but I'm not sure where it would come from. Anyone know what I'm doing wrong?

$\endgroup$
3
  • $\begingroup$ Hello there, and welcome to the Physics Stack Exchange! Homework and "check my work" questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. Please read this post on asking homework questions and this post for "check my work" questions. $\endgroup$ Jun 9, 2021 at 16:14
  • $\begingroup$ Are you trying to get a horizontal component of gravity? $\endgroup$
    – DJohnM
    Jun 9, 2021 at 16:36
  • $\begingroup$ @DJohnM if you're referring to equation (2), I'm trying to get the component of gravity normal to the tension in the string $\endgroup$
    – VanceR
    Jun 9, 2021 at 17:28

2 Answers 2

2
$\begingroup$

Forces on the point charge.

You have not considered the tension in your working. It's usually good practice to draw the force vectors out - the diagram shows the tension, $T$, electrostatic repulsion force, $F_e$, and gravitational force, $F_g$, acting on the point charge.

We see that for equilibrium, the vertical component of the tension, $T\cos{\theta}$ must balance out with the gravitational force/weight; and the horizontal component of the tension, $T\sin{\theta}$ must balance out with the electrostatic repulsion force. Hence, all your equations won't give you the correct result since you have neglected the tension. The equations that describe the equilibrium are instead:

$T \cos{\theta} = mg$ for vertical equilibrium and

$T\sin{\theta} = \frac{q^2}{4 \pi \epsilon_0 (2L\sin{\theta})^2} = \frac{k_e q^2}{(2L\sin{\theta})^2}$ for horizontal equilibrium.

Dividing both equations to get rid of the tension, $T$, and re-arranging should get you the desired result.

$\endgroup$
0
$\begingroup$

This problem should be address via potential energy vs $\theta$:

$$U(\theta) = -2Lmg\cos\theta +\frac{k_eq^2}{2L\sin\theta}$$

and solving for:

$$\frac{dU}{d\theta} =2Lmg\sin\theta -\frac{k_eq^2}{2L}\csc\theta\cot\theta=0$$

$$q^2 = (2L)^2\frac{mg}{k_e} \frac{sin^3\theta}{\cos\theta}$$

$$ q = \pm 2L\sin{\theta}\sqrt{\frac{mg\tan\theta}{k_e}}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.