Question

PLZZ solve the problem​

Answer 1

Answer:

First of all find out the electric field due to a quadrant i.e. one fourth of a circle . So let us take the quadrant with charge per unit length is 2lambda ..... From horizontal direction make an angle theta and then take an angle d(theta) to cut a small elemental charge dq which is equal to lambda × R × d(theta)

Where R x d(theta) is the elemental length that we have cut down which contains dq charge

Now find the electric field due to the elemental length by breaking the dE into components in x and y axis then integrate it as shown in the image ...... So now you have found the electric field components due to one quadrant in x and y direction..

Now find out electric field due to all the quadrants and add them vectorially..

Then finally we get

E(net) = 2k(lambda) / R in negative x direction i.e. minus i cap

electric field due to

for the first quadrant

in negative x axis is 2k(lambda)/R

in negative y axis is 2k(lambda)/R

for second quadrant

in negative x axis is 2k(lambda)/R

in positive y axis is 2k(lamba)/R

for third quadrant is

in positive y axis is k(lamba)/R

in positive x axis is k(lambda)/R

for fourth quadrant is

in negative y axis is k(lambda)/R

in positive x axis is k(lambda)/R

add all of them vectorially i.e. if Electric field in positive direction then add it with positive sign and if Electric field is in negative x axis then add it with a negative sign .....

hope it will help you