Question

Please explain part A B C and D of this problem

Constants A ring-shaped conductor with radius a 2.20 cm has a Part A total positive charge Q-0.129 nC uniformly distributed around it.(Figure 1) What is the magnitude of the electric field at point P, which is on the positive x-axis at z = 44.0 cm ? You may want to review (Pages 699- 704) Figure < 1of1 > N/C Submit Previous Answers Request Answer ds X Incorrect; Try Again; 2 attempts remaining -Part B What is the direction of the electric field at point P? *x-direction

Answer 1

A) In order to calculate the electric field due to a ring of charge Q, we consider a small part of ring of length do, and then find the electric field due to it at point P.

This electric field is given by dE and can be broken into components along x axis and y axis.

In this case we can see that the components along the y axis get cancelled because due to the half rings the y components of the electric force act in the opposite direction.

Therefore the net electric field = integration of dE in the x direction for part of ring dl varying from 0 to 2πa with a charge dQ, where dQ=Qdl/2πa

B) The direction of electric field at point P I towards the +x axis as we can see in the diagram.

C) Electric force on a charge due to an Electric field = electric field x charge.

D) since the ring is +vely charged, and the charge exerting the force is negative, therefore itlit be attractive in nature, and since the charge is negative itlitbe towards the charge, i.e. +x axis.