# Homework Help: 1.) Net electric field and 2.) Maximum electric field from centre of a ring

1. May 13, 2007

### irnubcake

1. Problem1 statement, all variables and given/known data

In the figure particle 1 of charge q1 = -8.13q and particle 2 of charge q2 = +3.63q are fixed to an x axis. As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero?

http://img201.imageshack.us/img201/4669/netfieldzeromy7.gif [Broken]

2. Relevant equations

Electric Field = k|q| / r²

The attempt at a solution

Since it's a point between the two charged particles, let x = the distance between q1 and that point, so the distance between the point and q2 = L - x

http://img157.imageshack.us/img157/990/netfieldzero2xj8.gif [Broken]

The net electric field = 0,
so -(Electric field due to q1) + (Electric field due to q2) = 0
E1 is negative because q1 is negative, E2 vice versa.

-( k|-8.13q| / x² ) + ( k|3.63q| / (L - x)² ) = 0
Cancelling, etc gives:
4.5x² - 16.26Lx + 8.13L² = 0
quadratic formula gives: x = 0.6L or 3.01L

I got it wrong, I'm also confused about the sign of the electric fields but the quadratic formula won't produce a real solution if both fields were negative or positive.

1. Problem2 statement, all variables and given/known data

Charge is uniformly distributed around a ring of radius R = 2.41 cm and the resulting electric field is measured along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's center is E maximum?

2. Relevant equations

Electric Field of Ring = (kqx) / ( (x² + k²) ^ (3/2) )

The attempt at a solution

I attempted to differentiate the equation with respect to x and put E' = 0, but I have two unknowns, x and q, the solution is a real number with no variables.

Last edited by a moderator: May 2, 2017
2. May 13, 2007

### Staff: Mentor

Problem1

Why do you assume that the point in question is between the two particles? Is that even possible? There are three regions to consider: left of q1, between q1 & q2, right of q2. Step one is to figure out which region is a candidate for containing the zero field point. (In your diagram, draw the directions of the fields from each charge.)

3. May 13, 2007

### Staff: Mentor

Problem2

Electric Field of Ring = (kqx) / ( (x² + R²) ^ (3/2) )

You might not know the value of q, but it is a constant. It will drop out of your final equation. Try it!

4. May 13, 2007

### irnubcake

Ah for the second problem, it ended up as x = R / sqrt(2)
x = 2.41 cm / sqrt(2) = 1.7041 cm

As for the first: http://img515.imageshack.us/img515/4338/netfieldzero3ir5.gif [Broken]

So I suppose a net field of zero can't be between them. That leaves the left and right regions. Hmmm so maybe the distances: L + x and x. But I'm confused about the signs of the electric fields. Suppose the point is to the right of q2 would E1 still be negative and E2 positive?

Last edited by a moderator: May 2, 2017
5. May 13, 2007

### Staff: Mentor

Yes. Think of it this way: Since q1 is negative, E1 always points towards q1; so, for x > 0, E1 points to the left and is thus negative. And for x < 0, E1 points to the right and is positive.

Similarly, E2 always points away from q2, a positive charge. For x > L, E2 is towards the right and thus positive.

6. Sep 18, 2008

### s7b

I don't understant how to take the derivative of the second question. I end up with a huge mess....help?

7. Sep 18, 2008

### Staff: Mentor

Just use the quotient or product rule with the chain rule.

8. Aug 27, 2009

### kaymant

For the second one u need not use derivatives. We need the max of
$$\dfrac{x}{(x^2+R^2)^{3/2}}$$
Since the expression is odd w.r.t. $$x$$, we may as well assume $$x>0$$. Then the given expression
$$=\dfrac{1}{\left(\dfrac{x^2}{x^{2/3}}+ \dfrac{R^2}{x^{2/3}}\right)^{3/2}}= \dfrac{1}{\left(x^{4/3}+ \dfrac{R^2}{x^{2/3}}\right)^{3/2}}$$
Hence, the original expression is going to be maximum where the quantity inside the radical in the denominator, i.e. the quantity $$x^{4/3}+ \dfrac{R^2}{x^{2/3}}$$ of the last expression is minimum. But by A.M -G.M inequality we get
$$x^{4/3}+ \dfrac{R^2}{x^{2/3}} = x^{4/3}+\dfrac{R^2}{2x^{2/3}}+\dfrac{R^2}{2x^{2/3}} \geq 3\sqrt[3]{x^{4/3}\cdot \dfrac{R^2}{2x^{2/3}}\cdot \dfrac{R^2}{2x^{2/3}}}$$
The point is that the condition for equality holds when
$$x^{4/3} = \dfrac{R^2}{2x^{2/3}}\quad \Rightarrow\ \boxed{x=\dfrac{R}{\sqrt{2}}}$$