3 identical point charges of magnitude 6nC are placed at the 3 corners of a square 40mm on a side. What is the magnitude and direction of the electric field due to the 3 charges at the vacant corner.

Question

Electric Fields in Physics

Answer 1

Let

q = 6 nC be the size of each charge,

s = 40 mm = 0.04 m be the size of the square,

k = 9×10⁹ Nm²/C² be Coulomb's constant.

The vacant corner is affected by two charges at distance s.

Each creates an electric field of magnitude

E = kq/s²

Those two field vectors are perpendicular to each other, so their vector sum will have the magnitude √2kq/s².

This vector sum points in the same direction as the field due to the corner

across the diagonal; therefore their magnitudes can be added.

The magnitude of this third vector is kq/(s√2)².

Thus the resulting field points in the direction away from the diagonal, and has the magnitude

√2kq/s² + kq/2s² = kq(2√2 + 1)/2s²

Substituting actual numbers

9×10⁹×6×10^-9×(2√2 + 1)/(2×0.04²) = 64604.7 N/C