Question

1. Calculate the magnitude of the electric field at the center of a square 42.5cm on a side if one corner is occupied by a −38.9μC charge and the other three are occupied by −26.5μC charges. Choose the correct direction of the electric field at the center of the square.

2. Choose the correct direction of the electric field at the center of the square.

A. toward to −38.9μC charge

B. perpendicular to the diagonal passing through −38.9μC

C. charge outward of −38.9μC charge

Answer 1

Electric field at the center due to the charge 1
$E_{1} = \frac{kq_{1}}{r^{2}}$
$E_{1} = \frac{9*10^{9}*38.9*10^{-6}}{0.30^{2}}$
E₁ = 3.89 N/C
Now the electric field due to the charge 2
$E_{2} = \frac{kq_{2}}{r^{2}}$
$E_{2} = \frac{9*10^{9}*26.5*10^{-6}}{0.30^{2}}$
E₂ = 2.65 N/C
Electric field due to the charge 3 and 4 will be same as E₂ due to symmetry , therefore
E₂ = E₃ =E₄
E₂ and E₄ will cancelled each other
Hence the net field at the center will be
E = E₁ - E₃ = 3.89 - 2.65 = 1.24 N/C
(2)
the direction will be toward - 38.9 $\mu$C
Hence the correct option is A. i.e toward to - 38.9 $\mu$C