Question

A uniformly charged insulating rod of length 13.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of -6.50

Answer 1

Draw a horizontal line through O to intersect the rod at P.

Now consider a small part of the rod such that the perpendicular from O to that part is inclined at an angle ? with OP. Let this part subtend an angle d? at the center.

So length of this part = R d?

Charge of this part = R d? * Q / l
where Q and l are the charge & length of the entire rod.

Electric field due to this charge
dE = k*(R d? * Q / l) / R^2 = k Q d? / Rl

The direction of this electric field is along the radius joining that part to the center O.

Now consider another part below OP such that the radius through this part is also inclined to OP at an angle ?.

The electric field exerted by this part will also be dE.

Now the vertical components of fields due to these two mirror elements cancel each other and the net field is along OP.

Similarly, the vertical field due to any part will be canceled by that due to its mirror element.

So net field at O will be the sum of components of fields due to all parts along OP.

Component of field due to a part along OP = dE cos? = k Q cos? d? / Rl

For E(net) integrate from ? = -?/2 to ? = ?/2.

E(net) = 2 k Q / R l

Also l = ?R

So
E(net) = 2 ? k Q / l^2

Substitute:
k = 9 x 10^9
Q = 6.5 x 10^-6 (Take it to be positive to calculate the magnitude of E. Then calculate direction separately)
l = 0.13 m

E (net) = 2174948.76 N/C

Direction is along OP.