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Two 10-cm-diameter charged rings face each other, 15.0 cm apart. Both rings are charged to + 30.0 nC . What is the electric field strength at the midpoint between the two rings and at the center of the left ring
The Answer
a) Zero. If you put a test charge there, the force from the left ring exactly cancels the force from the right ring - due to the symmetry.
b) The left ring itself has no contribution to the total field at its centre, same argument as part a).
All the field is from the right hand ring. Really you need calculus to do this, but I'll try and explain without it.
Imagine the axes of the rings are horizontal. Call the centre of the left ring 'P'.
Each element (small piece) of the right ring is a distance d from P where
d = sqrt(20^2+5^2) = 20.6cm = 0.206m
The angle subtended by the element at P = tan-1(5/20) = 14deg
If each element carries charge q then the field from it at P is kq/(0.206)^2 (where k=Coulomb's constant, 9x10^9units)
The horizontal component of this field contribution at P is kqcos14/(0.206)^2
Not that all the field components perpendicular to the axis will cancel (symmetry again). We just have to sum the horizontal (axial) field components.
The total field = the sum of the fields from all the elements. If the total charge is Q, this is just
kQcos14/(0.206)^2
which is the answer - you can do it on your calculator!
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