Equilateral triangle charged rods

[oreosama]

Homework Statement

Three rods each of uniform charge Q and length a are formed into an equilateral triangle. Given[Q,a,q] determine:
1. The voltage at the center of the triangle.
2. If a charge -q is placed at the center of the triangle determine potential energy.
3. Determine the initial speed of the -q in order that it exits the system and never returns.

Homework Equations

$$V = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r}$$

$$U = Vq$$

$$K_0 + U_0 = K + U$$

The Attempt at a Solution

http://i.imgur.com/7IZ3PbP.png

I take 1 half of bottom rod and intend to integrate for V$$dq = Q \frac{dx}{a}$$
$$r = \sqrt{y^2 + (\frac{a}{2} - x)^2}$$

using the V formula above this integrates(from 0 to a/2) to$$V = \frac{Q}{4 \pi \epsilon_0}[\ln{\frac{a+\sqrt{\frac{a^2}{4} + y^2}}{y}}]$$

since this is 1/6th of the potential and they are additive, multiply 6 for total v at centeris this correct solution? U is simple enough to do, simply multiply by -q. What is the methodology for solving part 3?

 

[voko]

You might simplify things significantly if you express $y$ via $a$.

 

[oreosama]

$$y = \sqrt{\frac{3}{4}} a$$

ln term collapses to
$$ln{\frac{4}{3}}$$

Is this correct? What about part 3? I only know of 1/2mv^2 which I assume was at least greater than potential it would get out.. but I have no m given.

 

[voko]

"Never returns" means that it goes infinitely far away, and the velocity "at the infinity" is at least zero.

With this in mind, apply conservation of energy.

 

[Nickie]

I would like to commend you for your attempt at solving this problem and using the appropriate equations. However, I do have some comments and suggestions for improvement.

Firstly, your solution for the voltage at the center of the triangle is incorrect. The equation you used, V = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r}, is the correct formula for voltage due to a continuous charge distribution. However, in this problem, we are dealing with point charges (the three rods) arranged in a specific geometry. Therefore, you cannot simply integrate over one half of the bottom rod. Instead, you need to calculate the voltage due to each individual rod and then add them together.

Secondly, your integration limits are incorrect. The limits should be from 0 to a, as you are integrating over the entire length of the rod, not just half of it.

Thirdly, you did not specify the distance from the center of the triangle to the center of the bottom rod in your integration. This distance, y, should be included in your equation for r.

Finally, for part 3, you can use the conservation of energy equation, K_0 + U_0 = K + U, to determine the initial speed of the -q charge. However, you need to specify the initial conditions, such as the initial potential energy and the distance from the center of the triangle to the center of the -q charge.

In conclusion, while your attempt at solving the problem is commendable, your solution is not entirely correct. I suggest reviewing the concepts of voltage and potential energy due to point charges and then attempting the problem again. It may also be helpful to consult a textbook or seek assistance from a teacher or tutor.