Finding the Electric Field at the Midpoint of two rings

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SUMMARY

The electric field at the midpoint between two charged rings, each with a diameter of 10 cm and separated by 20 cm, can be calculated using the principle of superposition. The left ring has a charge of -22 nC and the right ring has a charge of +22 nC. The correct formula for the electric field due to a ring is different from that of a point charge; thus, the total electric field is derived from the axial field formula for a ring. The final calculated electric field at the midpoint is 395,560 N/C.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with the principle of superposition in electrostatics
  • Knowledge of the axial electric field formula for charged rings
  • Basic calculus for integrating charge distributions
NEXT STEPS
  • Study the axial electric field formula for a charged ring
  • Learn about the principle of superposition in electrostatics
  • Explore charge distribution and its effects on electric fields
  • Practice problems involving multiple charged objects and their electric fields
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Students studying electromagnetism, physics educators, and anyone interested in electrostatic field calculations involving charged rings.

arileah
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Homework Statement


Hello,

Two 10-cm-diameter charged rings face each other, 20 cm apart. The left ring is charged to -22 nC and the right ring is charged to +22 nC . What is the magnitude of the electric field E⃗ at the midpoint between the two rings?

Homework Equations


[/B]
E = Kq/r^2
K = 8.99 * 10^9

The Attempt at a Solution



Find the electric field at the midpoint caused by each individual plate, then using the principle of superposition to add them.

E(tot) = |E(1)| + |E(2)|

|E(1)| = K (22nC)/0.1^2
|E(2)| = K (22nc)/0.1^2

|E(1)| + |E(2)| = E {since E(1) and E(2) are the same}

E(tot) = E = 2 * K (22nC)/0.1^2 = 395560 N/C

However, the answer I got is wrong. I have also tried E = 0 N/C in case I was not supposed to take the magnitudes. This is also incorrect.

Could anyone lend me a hand?
 
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arileah said:
K (22nC)/0.1^2
That would be the field 10cm from a point charge of 22nC.
The charge here is distributed around a ring (not a plate). At the centre of each ring there is no field induced by that ring since the fields due to the charges around the ring point in all directions through the centre and, by symmetry, cancel.
Consider a small portion of one ring, length rdθ, carrying charge q, where r is the ring's radius. And consider a point P distance x from the centre of the ring, along the axis of the ring. How far is P from the charge q? What is the strength of the field at P due to the charge q? In what direction does that field point?

Alternatively, you may have already been taught a formula for the axial field due to one ring, in which case all you have to do is double it for the second ring.
 
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Thank you! I got it now :)
 

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