Physics - Electrostatic Force Question

[VitaX]

Homework Statement

Homework Equations

F = k*Q1*Q2/r^2

The Attempt at a Solution

My work is E(x) = (k*Q1)/(x^2) - (k*Q2)/((L-x)^2)

Find derivative and set E(x) = 0. But I get a non real answer. Am I doing this completely wrong here? I was going to solve for x here.

 

[gneill]

It could be that the problem is with the small numbers that are in play. How are you going about solving the resulting cubic? Using a machine method or solver?

You may find it convenient to discard all unnecessary constants, units, and orders of magnitude that are simply scaling factors. Set Q1 = 1, Q2 = 26.7, L = 6.92. Minimize:

f(x) = Q1/x² + Q2/(L-x)²

 

[SammyS]

Q₂ = ‒ 26.7e .

plug in Q₁ and Q₂ into your equation, E(x) = (k·Q₁)/(x²) - (k·Q₂)/((L-x)²).

That should give a real answer.

 

[VitaX]

So you're saying disregard using -1.6E-19 C in place of e? Also which of your equations should I use to solve for x than find minimum F value?

 

[gneill]

Scaling factors like k and e don't affect the shape of the curve, or where the minimum occurs. Accounting for the sign of the Q2 charge, the equation I wrote above in post #2 should lead to your minimum. Interpret the result as centimeters.

 

[VitaX]

Scaling factors like k and e don't affect the shape of the curve, or where the minimum occurs. Accounting for the sign of the Q2 charge, the equation I wrote above in post #2 should lead to your minimum. Interpret the result as centimeters.

I'm still a bit confused though on how to find the x value. Are you saying I should take the derivative of your equation with respect to x and set that equal to zero and solve for x? Then from one of the values of x, plug that into the original equation for F and that gives my minimum?

 

[gneill]

I'm still a bit confused though on how to find the x value. Are you saying I should take the derivative of your equation with respect to x and set that equal to zero and solve for x? Then from one of the values of x, plug that into the original equation for F and that gives my minimum?

Sure. You don't even have to plug it back into your original formula, since you only need the value of x (the position of where the minimum occurs) Only one of the values of x returned by solving for x will be 'physically realistic'.

 

[VitaX]

Sure. You don't even have to plug it back into your original formula, since you only need the value of x (the position of where the minimum occurs) Only one of the values of x returned by solving for x will be 'physically realistic'.

But won't I have to plug it back into the original in order to find that minimum magnitude for part b?

 

[gneill]

But won't I have to plug it back into the original in order to find that minimum magnitude for part b?

Hey, I didn't look that far ahead! So, yes. Plug away!

 

[VitaX]

I did this:

Derivative of the function F = Q1/x^2 + Q2/(L-x)^2 is dF/Dx = -2Q1/x^3 + 2Q2/(L-x)^3 = 0

Q2/(L-x)^3 = Q1/x^3 -----> Q2*x^3 = Q1(L-x)^3
Q2^(1/3)*x = Q1^(1/3)*(L-x) -----> 2.9888x = .0692 - x -----> x = .0173 m or 1.735 cm

F = 1/.0173^2 + 26.7/(.0692-.0173)^2 = 13253.59 N

I used L = .0692 m
Q1 = 1
Q2 = 26.7

Is this right? At this point I'm blank as to whether or not it is because I didn't utilize -1.6E-19 C in place of e, but rather ignored them. I guess it was alright to do that to find x in meters. But do I utilize those values now for finding the minimum magnitude F?

Edit: Now I'm really lost since I just noticed it gives Q2 as -26.7 and not positive.

 

[gneill]

I did this:

Derivative of the function F = Q1/x^2 + Q2/(L-x)^2 is dF/Dx = -2Q1/x^3 + 2Q2/(L-x)^3 = 0

Q2/(L-x)^3 = Q1/x^3 -----> Q2*x^3 = Q1(L-x)^3
Q2^(1/3)*x = Q1^(1/3)*(L-x) -----> 2.9888x = .0692 - x -----> x = .0173 m or 1.735 cm

F = 1/.0173^2 + 26.7/(.0692-.0173)^2 = 13253.59 N

I used L = .0692 m
Q1 = 1
Q2 = 26.7

Is this right? At this point I'm blank as to whether or not it is because I didn't utilize -1.6E-19 C in place of e, but rather ignored them. I guess it was alright to do that to find x in meters. But do I utilize those values now for finding the minimum magnitude F?

Edit: Now I'm really lost since I just noticed it gives Q2 as -26.7 and not positive.

You've done fine. Scaling factors that apply to all the terms of what you're minimizing or maximizing can be ignored while doing the minimizing/maximizing. Of course you have to use them in calculating the actual value of the function. Finding a minimum or maximum didn't change the physics!

Regarding Q2 being a negative charge, that was taken into account when the formula for Fnet was written; the net force from both charges Q1 and Q2 on charge Q3 is towards +x.

 

[SammyS]

I did this:

Derivative of the function F = Q1/x^2 + Q2/(L-x)^2 is dF/Dx = -2Q1/x^3 + 2Q2/(L-x)^3 = 0

Q2/(L-x)^3 = Q1/x^3 -----> Q2*x^3 = Q1(L-x)^3
Q2^(1/3)*x = Q1^(1/3)*(L-x) -----> 2.9888x = .0692 - x -----> x = .0173 m or 1.735 cm

F = 1/.0173^2 + 26.7/(.0692-.0173)^2 = 13253.59 N

I used L = .0692 m
Q1 = 1
Q2 = 26.7

Is this right? At this point I'm blank as to whether or not it is because I didn't utilize -1.6E-19 C in place of e, but rather ignored them. I guess it was alright to do that to find x in meters. But do I utilize those values now for finding the minimum magnitude F?

Edit: Now I'm really lost since I just noticed it gives Q2 as -26.7 and not positive.

!.735 cm is correct.

You used ‒Q₂ rather than Q₂ to get that answer. If Q₂ -26.7 e, then ‒Q₂ = +26.7 e.

So, you're alright.

This would have come out cleaner if at the beginning, you had set:

E(x) = (k·e)/(x^2) ‒ (k·(‒26.7e))/((L‒x)^2) = (k·e)[1/(x²) + 26.7/(L‒x)²) ]

Then d/dx(E(x)) = 0  →  (k·e)d/dx[1/(x²) + 26.7/(L‒x)²)] = 0 and you can divide by k·e to get rid of it when finding the minimum.

 

[VitaX]

I'm a bit confused at what the e in the intial question stand for. They aren't electron that's for sure because I get a negative Force value when I assume they are and plug it into the original equation.

 

[gneill]

I'm a bit confused at what the e in the intial question stand for. They aren't electron that's for sure because I get a negative Force value when I assume they are and plug it into the original equation.

I believe that the e is meant to represent the fundamental charge. That is why they specified the sign of the charge separately.

The fundamental charge is the magnitude (positive value) of the charge on the electron (or proton).

 

[SammyS]

It is weird, however, to have a charge of magnitude 26.7 e.

 

[VitaX]

I just used 1.6E-19 C as e. My friend said it means fundamental charge, which coincidentally is the charge of a proton. I took the e as meaning electron but he said it doesn't. I believe that's what gneill stated as well so I get a positive Force now.