Four identical charges of 2.0 × 10^ -6 C each are secured at the...

Let the four charges $q = 2.0 \times 10^{-6} \text{ C}$ be placed at the corners A, B, C, and D of a square of side $a = 0.05 \text{ m}$.

We calculate the net force on the charge at corner D due to the charges at A, B, and C.

The force due to the charge at A is directed along AD:

$F_A = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{a^2} = \dfrac{9 \times 10^9 \times (2.0 \times 10^{-6})^2}{(0.05)^2} = 14.4 \text{ N}$

The force due to the charge at C is directed along CD:

$F_C = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{a^2} = 14.4 \text{ N}$

The force due to the charge at B is directed along the diagonal BD:

$F_B = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{(\sqrt{2}a)^2} = \dfrac{F_A}{2} = 7.2 \text{ N}$

The forces $F_A$ and $F_C$ are perpendicular to each other. Their resultant $F_{AC}$ is directed along the diagonal BD:

$F_{AC} = \sqrt{F_A^2 + F_C^2} = 14.4\sqrt{2} \approx 20.36 \text{ N}$

Since $F_{AC}$ and $F_B$ are in the same direction, the net force on the charge at D is:

$F_{net} = F_{AC} + F_B = 20.36 + 7.2 = 27.56 \text{ N} \approx 27.5 \text{ N}$

The resultant force is directed along the diagonal, which makes an angle of 45^{}^{{}^{\circ}} with the extended sides of the square.