Question

Electricity and Magnetism Physics - 1st year university level.

I am clueless on how to do this problem.

3) Two "quarter squares" wires with side length a (the full square would have a side length of 2a), are located in the top half of the plane, centred on the origin. The left "quarter square" is uniformly charged, with a total charge-a, while the right "quarter square" is uniformly charged with a total charge +Q. Calculate the electric field (magnitude and direction) at the origin created by this distribution of charge. Hint: you can use superposition of line of charges. 2a

Answer 1

Total length of the arm having +ve charge = a + 2 a = 3 a So linear charge density for positive side lambda_1 = Q/3 a Similarly linear charge density for negative side, lambda_2 = -Q/3 a Consider an element of length d y on right vertical side then total charge on the element d q = lambda_1 times d y = Q/3 a d y The electric filed d E^rightarrow due to charge element is sown in the figure. The vertical and horizontal components of d E^rightarrow (i.e d E cos theta & d E sin theta) are also shown in figure. If we take a symmetrical charge element on left -ve vertical side A B. Then due to symmetry electric field d E^rightarrow produced by this element will be of same magnitude as d E^rightarrow. If we resolve d E^rightarrow into vertical and horizontal components then we see that d E sin theta & d E� sin theta