Calculate the magnitude of the electric field at one corner of a square 1.82 m on a side if the other three corners are occupied by 4.75×10−6 C charges.
Solution)
Here, we have F1= F3= kq/a^2=(9*10^9)*(4.75*10^-6)/1.82^2 =1.29*10^4 N/C
F2= kq/2a^2=6.45*10^3 N/C
Now, Fr= sqrt(F1^2 + F3^2) + F2
Fr= 2.46*10^4 N/C
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