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Uniformly charged ring on the axis

  • Thread starter SlyCedix
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  • #1
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Homework Statement


The electric field on the axis of a uniformly charged ring has magnitude 360 kN/C at a point 6.2 cm from the ring center. The magnitude 19 cm from the center is 150 kN/C ; in both cases the field points away from the ring.

Homework Equations


1. Find the ring's radius.
2. Find the ring's charge.

The Attempt at a Solution


E=kxQ/(x2 + a2)3/2
360000 = kQ(.062)/(.0622+a2)3/2
150000 = kQ(.19)/(.192+a2)3/2

360/150 = (.062/.19) * ((.0622+a2)/(.192+a2))3/2

(360/150 * .19/.062)2/3 = (.0622+a2)/(.192+a2)

(360/150 * .19/.062)2/3 * (.192+a2)= .0622+a2

(360/150 * .19/.062)2/3 * (.192) - .0622 = a2 - (360/150 * .19/.062)2/3 * a2)

-0.00247871714 = -2.7819469745 a^2

a = sqrt(0.00247871714/2.7819469) = .030 m = 3.0 cm

Which turned out to be incorrect, and with an incorrect radius, I've got nothing to go off of for charge
 
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Answers and Replies

  • #2
SammyS
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Homework Statement


The electric field on the axis of a uniformly charged ring has magnitude 360 kN/C at a point 6.2 cm from the ring center. The magnitude 19 cm from the center is 150 kN/C ; in both cases the field points away from the ring.

Homework Equations


1. Find the ring's radius.
2. Find the ring's charge.

The Attempt at a Solution


E=kxQ/(x2 + a2)3/2
360000 = kQ(.062)/(.0622+a2)3/2
150000 = kQ(.19)/(.0192+a2)3/2

360/150 = (.062/.19) * ((.0622+a2)/(.0192+a2))3/2

(360/150 * .19/.062)2/3 = (.0622+a2)/(.0192+a2)

(360/150 * .19/.062)2/3 * (.0192+a2)= .0622+a2

(360/150 * .19/.062)2/3 * (.0192) - .0622 = a2 - (360/150 * .19/.062)2/3 * a2)

-0.00247871714 = -2.7819469745 a^2

a = sqrt(0.00247871714/2.7819469) = .030 m = 3.0 cm

Which turned out to be incorrect, and with an incorrect radius, I've got nothing to go off of for charge
Hello @SlyCedix . Welcome to PF.


It's just a little Algebra mistake.

What do you get when you divide ##\displaystyle \ \frac{kQ (0.062)} {(0.062^2+a^2)^{3/2}} \ ## by ##\displaystyle \ \frac{kQ (0.19)} {(0.19^2+a^2)^{3/2}} \ ##?
 
  • #3
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Wow, I cannot believe I missed that.

So the revised eq would be:

$$\frac{360}{150} = \frac{.062}{.19} * (\frac{(.19^2+a^2)}{(.062^2+a^2)})^\frac{3}{2}$$

$$(\frac{360}{150} * .\frac{.062}{.19})^\frac{2}{3} = \frac{(.19^2+a^2)}{(.062^2+a^2)}$$

$$(\frac{360}{150} * \frac{.062}{.19})^\frac{2}{3}* (.062^2+a^2)= .19^2+a^2$$

$$(\frac{360}{150} * \frac{.062}{.19})^\frac{2}{3} * .062^2 - .19^2 = a^2 - (\frac{360}{150} * \frac{.062}{.19})^\frac{2}{3} a^2$$

$$0.033212 = -2.7819469745 a^2$$

~~But that would mean a is non real, despite there being a real answer.~~

Scratch that, put it in my calculator wrong, it's actually

$$-0.02156 = -2.7819469745 a^2$$

Which is still wrong, but closer
 
  • #4
SammyS
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Wow, I cannot believe I missed that.

So the revised eq would be:

$$\frac{360}{150} = \frac{.062}{.19} * (\frac{(.19^2+a^2)}{(.062^2+a^2)})^\frac{3}{2}$$

$$(\frac{360}{150} * .\frac{.062}{.19})^\frac{2}{3} = \frac{(.19^2+a^2)}{(.062^2+a^2)}$$

$$(\frac{360}{150} * \frac{.062}{.19})^\frac{2}{3}* (.062^2+a^2)= .19^2+a^2$$

$$(\frac{360}{150} * \frac{.062}{.19})^\frac{2}{3} * .062^2 - .19^2 = a^2 - (\frac{360}{150} * \frac{.062}{.19})^\frac{2}{3} a^2$$

$$0.033212 = -2.7819469745 a^2$$

But that would mean a is non real, despite there being a real answer.
You made a similar mistake as previously, when going from the first to the second line .
 
  • #5
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That was it, thanks, answer was 8.8cm
 
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  • #6
SammyS
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That was it, thanks, answer was 8.8cm
Good.

What total charge did you get ?
 
  • #7
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800 nC
 
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