Problem 26.11 Part A Two 10-cm-diameter charged rings face each other, 30 cm apa

Question

Problem 26.11 Part A Two 10-cm-diameter charged rings face each other, 30 cm apart. The left ring is charged to -17 nC and the right ring is charged to +17 nC What is the magnitude of the electric field E at the midpoint between the two rings? Express your answer to two significant figures and include the appropriate units. E-Value Units Submit My Answers Give Up Part B What is the direction of the electric field E at the midpoint between the two rings? O To the left ring. Parallel to the plane of the rings. O To the right ring Submit My Answers Give Up

Explanation / Answer

Electric field due to a charged ring of radius 'a' and total charge 'Q' at a point on its axis at a distance 'x' from its centre is given by:

E = kQx/(x2+a2)3/2 (directed along the axial direction away from the ring for a positive test charge)

In this question,

Q1 = Q = 17*10-9 C, Q2 = -Q = -17*10-9 nC, a = 5 cm = 0.05 m, x = 15 cm = 0.15 m

and, we know, k = 8.99 * 109 N m2 C-1

Electric field due to ring 1 is,

E1 = kQ1x/(x2+a2)3/2 = [8.99*109]*[17*10-9]*0.15/(0.152 + 0.052)3/2 = 5.8 * 10-9 N/C (away from the ring, along the axis)

Electric field due to ring 2 is,

E2 = kQ2x/(x2+a2)3/2 = [8.99*109]*[17*10-9]*0.15/(0.152 + 0.052)3/2 = -5.8 * 10-9 N/C (towards the ring, along the axis)

Part A: So, magnitude of combined electric field, |Etotal| = |E1| + |E2| = (5.8 * 10-9 + 5.8 * 10-9) N/C = 1.16 * 10-8 N/C

Part B: Direction of electric field is along the common axis of the rings towards the negatively charged ring.

Part C: Magnitude of force on the -1.0 nC charge placed at midpoint, |F->| = |-1*10-9| |Etotal| = 1.16 * 10-17 N

Part D: The direction of the force is along the common axis of the rings towards the positive charged ring. (The direction of the force on a negative point charge is opposite to the direction of the electric field at that point.)

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