thefour particles form a square of edgelengtha5.00cm and have chargesq110.0nC,q220.0nC,q320.0nC,and q410.0nC.In unit-vector notation,what net electric fielddo the particles produce at the square’scenter?

r1 = (a/2)x - (a/2)y ; as the center is located in + a/2 distance from 1 in x and -a/2 distance from 1 in y

here, x and y are units vectors along x and y direction respectively.

similarly,

r2 = -(a/2)x + -(a/2)y

r3 = -(a/2)x + (a/2)y

r4 = (a/2)x + (a/2)y

now, in unit vector form

r1 = r1 / |r1|

so, for 1 

r1 = r1 / [(a/2)2 + (a/2)2]1/2

and thus, for rest

r2 = r2 / [(a/2)2 + (a/2)2]1/2

r3 = r3 / [(a/2)2 + (a/2)2]1/2

r4 = r4 / [(a/2)2 + (a/2)2]1/2

so, we get

r1 = (1/2).(x - y)

r2 = (1/2).(-x - y)

r3 = (1/2).(-x + y)

r4 = (1/2).(x + y)

now, 

the net electric field at the center will be given as the sum of all the electric fields due to the four corner charges

so,

E = k(q1/r12).r1 +  k(q2/r22).r2 +  k(q3/r32).r3 +  k(q4/r42).r4

or by putting the value of the units vectors...

E = [ k(q1/r12) . ((1/2).(x - y)) ]  +  [ k(q2/r22) . ((1/2).(-x - y)) ]  +  [ k(q3/r32) . ((1/2).(-x + y)) ]  +  [ k(q4/r42) . ((1/2).(x + y)) ]

thus, by solving  further, we get

E = (k√2 / a2). [( q1 - q2 - q3 + q4 ) x + (- q1 - q2 - q3 + q4) y ]

now, we are given

q1 = +10 nC

q2 = -20 nC

q3 = +20 nC

q4 = -10 nC

and

k = 9x109 Nm2/C2

q = 0.05m

putting, the values, we get

E = [ (9x109 x √2) / 0.052 ]. [ (+10 - (-20) - 20 + (-10)) x + (-10 - (-20) - 20 + -(10)) y ]

thus by solving further, we get

E = 1.02 x 105 N/C y

the x-components cancel out..

Regards

Arun (askIITians forum expert)