calculate the work of electric field force...

Question

Positive charge Q is evenly distributed on a thin ring with a radius of R, the centre of which is point O. Axis OP is pendicular to the plane containing the thin ring. The length of OP is 2R. Taking a positive charge q from the point O to the point P very slowly, how much is the work of electric field forcel? (tips: calculate the work by W=Uq and you need to obtain the U by integral)

please help TT

Answer 1

Usually U_e is electrical potential energy and V is voltage. The equation is W = qΔV = ΔU. Since, we want the work of the field, it is the negative of the Work required against the field.

The V expression is obtained from dV = dq/(4πε₀r) (Coulomb's law) where r is the distance from the dq to the point on the z axis of interest.z = 0 to z=2R (z is coordinate along OP)

The dq is is Q/(2πR) Rdθ and r =sqrt(R²+ z²) so V = Q/(4πεsqrt(R²+z²)) + C (or k_eQ/sqrt(R²+x²) + C

The result means that the voltage is equivalent to that from a point charge sqrt(R²+z²) away

You only care about the difference in V so the W = -q(V(2R)-V(0)) = -(k_eQq/R)(1/sqrt(2) - 1) (Constant of integration cancels). Work is positive as the field act s to push the charge upwards.

Please consider a tutor. Take care.