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1.) Net electric field and 2.) Maximum electric field from centre of a ring 
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#1
May1307, 04:14 AM

P: 5

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/img201/4...eldzeromy7.gif 2. Relevant equations Electric Field = kq / 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/img157/9...ldzero2xj8.gif 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. ( k8.13q / x² ) + ( k3.63q / (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. 


#2
May1307, 05:34 AM

Mentor
P: 41,436




#3
May1307, 05:44 AM

Mentor
P: 41,436

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


#4
May1307, 07:40 AM

P: 5

1.) Net electric field and 2.) Maximum electric field from centre of a ring
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/img515/4...ldzero3ir5.gif 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? 


#5
May1307, 07:54 AM

Mentor
P: 41,436

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


#6
Sep1808, 05:10 PM

P: 26

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



#8
Aug2709, 09:13 PM

P: 9

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] 


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