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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)
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