Monday

November 24, 2014

November 24, 2014

Posted by **Sandra** on Monday, February 15, 2010 at 6:40am.

We are interested in the electric field a perpendicular distance z away from the center of the ring.

At what distance from the center of the ring does the electric field become maximum?

Hint: The field for a ring of charge is:

Ering = kQz/ (z^2+R^2)^3/2

I understand that we would need to differentiate this equation and solve for when it equals zero..but I'm having real hard time differentiating. Could someone help me out?

This is what I have so far...

E'ring = kQ (z/[(z^2 +R^2)^3/2])

= {{1/(z^2+R^2)^3/2}} - {{z(z^2+R^2)^1/2}} / {{(z^2+R^2)^3}}

This is what I got by using the product rule. But I don't know how to get rid of the constant R from the equations and or how to simplify in such a way that I would be able to solve for Z.

- Physics and Calculus -
**dongo**, Monday, February 15, 2010 at 6:44amFirstly, you are to use not the product rule, but the quotient rule, check out the related wikipedia article. Secondly, there is no necessity to "get rid" of R. Your solution can depend on R, for example. Just treat R as a constant (In fact it is, Radius (R) = 2.7 times 10^(-1) m)

- Physics and Calculus -
**bobpursley**, Monday, February 15, 2010 at 8:58amYou made a number of errors.

= {{1/(z^2+R^2)^3/2}} - {{2z^2}} / {{(z^2+R^2)^5/2}}

set it equal to zero. multiply both sides by (z^2+R^2)^3/2

0=1-2z/(z^2+R^2)

so 2z=z^2+R^2

z^2-2z+R^2=0

z=(2+-sqrt(4-4R))/2=1+-sqrt(1-R)

check my math.

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