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E field at a point between two rings

  • #1
1. Two 10-cm-diameter charged rings face each other, 19.0 cm apart. Both rings are charged to + 50.0 nC. What is the electric field strength at the center of the left ring?



2. E = q/(4*pi*epsilon*r2



3. Ok, the E field from the left ring is zero at this point due to cancellation/symmetry. Using r = 0.19 cm in the above equation, I obtain 1.25*104 N/C, but the correct answer is 1.13*104. What am I doing wrong?
 

Answers and Replies

  • #2
Jmf
49
0
Just geometry, I think. At the centre of the left ring, the charges on the right ring will not be at a distance of 19.0cm away, but rather (by pythagoras) at sqrt(19^2 + 5^2) cm away.
 
  • #3
collinsmark
Homework Helper
Gold Member
2,894
1,232
Hello Linus Pauling,

As Jmf pointed out, the effective distance is a little greater that 19 cm due to the Pythagorean theorem.

But there's more to it that just that. If you break up the right ring into small sections of length dl, you'll find that certain components of |E| cancel out with the components caused by the dl on the opposite side of the ring. In other words, only one component of the electric field caused by the right ring doesn't cancel.

In summary there's two things to consider. You need to use the Pythagorean theorem to find the magnitude of the distance. Secondly, you need to use some trigonometry to find the only component that doesn't cancel.
 

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