Calculate the magnitude of the electric field at one corner of a square 1.48 m on a side if the other three corners are occupied by 4.95×10−6 C charges.
According to an above figure, the electric field (E1) and (E2) which are equal in magnitude & add to a resulting field (E12) = 2 (E1) pointing northwest at -450.
Then, we have
Etotal = E12 + E3
Etotal = 2 E1 + E3
The electric field due to charge 1 which will be given as -
E1 = ke q / d2
The electric field due to charge 3 which will be given as -
E3 = ke q / (2 d)2
Therefore, magnitude of the electric field at one corner of a square which will be given as -
Etotal = 2 E1 + E3
Etotal = 2 (ke q / d2) + ke q / (2 d)2
Etotal = [2 + (1/2)] (ke q / d2)
where, q = charge on each corner = 4.95 x 10-6 C
d = side of a square = 1.48 m
ke = proportionality constant = 9 x 109 Nm2/C2
then, we get
Etotal = [(1.414) + (0.5)] [(9 x 109 Nm2/C2) (4.95 x 10-6 C) / (1.48 m)2]
Etotal = [(1.914) (20338.7 N/C)]
Etotal = 3.89 x 104 N/C
We know that, N/C = V/m
Finally, we have
Etotal = 38.9 kV/m
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