Two 10- c m -diameter charged rings face each other, 20 c m apart. The left ring

Question

Two 10-cm-diameter charged rings face each other, 20 cm apart. The left ring is charged to  - 20 nC and the right ring is charged to  + 20 nC .

What is the magnitude of the force F? on a -1.0 nC charge placed at the midpoint?

Explanation / Answer

electric force due to negatively charged ring = k*Q*q*x/[x^2+R^2]^(1.5) => F1 = 9*10^9*20*10^-9*10^-9*0.1/[0.1^2 + 0.05^2]^(1.5) = 1.29*10^-5 C This F1 force will be directed towards the centre of positively charged ring. The force F2 due to positively charged ring will also be same in magnitude & direction as F1. so, total force = 2*F1 = 2.58*10^-5 C (directed towards the centre of +vely charged ring)

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