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`        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?`
2 years ago

Arun
23742 Points
```							Dear Abdul 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 yhere, x and y are units vectors along x and y direction respectively.similarly,r2 = -(a/2)x + -(a/2)yr3 = -(a/2)x + (a/2)yr4 = (a/2)x + (a/2)ynow, in unit vector formr1 = r1 / |r1|so, for 1 r1 = r1 / [(a/2)2 + (a/2)2]1/2and thus, for restr2 = r2 / [(a/2)2 + (a/2)2]1/2r3 = r3 / [(a/2)2 + (a/2)2]1/2r4 = r4 / [(a/2)2 + (a/2)2]1/2so, we getr1 = (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 chargesso,E = k(q1/r12).r1 +  k(q2/r22).r2 +  k(q3/r32).r3 +  k(q4/r42).r4or 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 getE = (k√2 / a2). [( q1 - q2 - q3 + q4 ) x + (- q1 - q2 - q3 + q4) y ]now, we are givenq1 = +10 nCq2 = -20 nCq3 = +20 nCq4 = -10 nCandk = 9x109 Nm2/C2q = 0.05mputting, the values, we getE = [ (9x109 x √2) / 0.052 ]. [ (+10 - (-20) - 20 + (-10)) x + (-10 - (-20) - 20 + -(10)) y ]thus by solving further, we getE = 1.02 x 105 N/C ythe x-components cancel out.. RegardsArun (askIITians forum expert)
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2 years ago
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