PLZZ solve the problem
Answers
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