1.) Net electric field and 2.) Maximum electric field from centre of a ring

In summary, the maximum electric field strength occurs when the charge particles are located at the point in the center of the ring that is closest to them by a distance of 1.7041 cm.
  • #1
irnubcake
5
0
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]

Homework 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?

Homework 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.
 
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  • #2
Problem1

irnubcake said:
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
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
Problem2

irnubcake said:

Homework Equations



Electric Field of Ring = (kqx) / ( (x² + ) ^ (3/2) )
That equation should read:
Electric Field of Ring = (kqx) / ( (x² + R²) ^ (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.
You might not know the value of q, but it is a constant. It will drop out of your final equation. Try it!
 
  • #4
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?
 
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  • #5
irnubcake said:
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?
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
I don't understant how to take the derivative of the second question. I end up with a huge mess...help?
 
  • #7
Just use the quotient or product rule with the chain rule.
 
  • #8
For the second one u need not use derivatives. We need the max of
[tex]\dfrac{x}{(x^2+R^2)^{3/2}}[/tex]
Since the expression is odd w.r.t. [tex]x[/tex], we may as well assume [tex]x>0[/tex]. Then the given expression
[tex]=\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}}[/tex]
Hence, the original expression is going to be maximum where the quantity inside the radical in the denominator, i.e. the quantity [tex]x^{4/3}+ \dfrac{R^2}{x^{2/3}}[/tex] of the last expression is minimum. But by A.M -G.M inequality we get
[tex]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}}}[/tex]
The point is that the condition for equality holds when
[tex]x^{4/3} = \dfrac{R^2}{2x^{2/3}}\quad \Rightarrow\ \boxed{x=\dfrac{R}{\sqrt{2}}}[/tex]
 

1. What is net electric field?

The net electric field is the combined effect of all the individual electric fields present in a given region. It is a vector quantity that describes the strength and direction of the overall electric field at a specific point.

2. How is net electric field calculated?

The net electric field can be calculated by vector addition of all the individual electric fields present in a given region. The magnitude and direction of each electric field is taken into account to determine the overall net electric field at a specific point.

3. What is the maximum electric field from the center of a ring?

The maximum electric field from the center of a ring is the highest value of electric field that can be measured at any point on the circumference of the ring. It is directly proportional to the charge and inversely proportional to the distance from the center of the ring.

4. How is the maximum electric field from the center of a ring calculated?

The maximum electric field from the center of a ring can be calculated using the formula E = kQ/r, where E is the electric field, k is the Coulomb's constant, Q is the charge on the ring, and r is the distance from the center of the ring to the point where the electric field is being measured.

5. What factors affect the maximum electric field from the center of a ring?

The maximum electric field from the center of a ring is affected by the charge on the ring, the distance from the center of the ring, and the medium in which the ring is placed. It is also affected by the presence of other charges or conductors in the vicinity of the ring.

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