Q.Four charges of $+6\times10^{-8}$ C each are placed at the corners of a square whose sides are 3 cm each. Calculate the resultant force on each charge and show its direction on a diagram drawn to scale.

[!TLDR] Adding the two equal side forces (perpendicular) and the diagonal force on a square-corner charge gives a resultant of $6.89\times10^{-2}$ N, directed straight out along that charge's own diagonal. [!ANSWER] Resultant force on each corner charge $\approx6.89\times10^{-2}$ N, directed radially outward along the square's diagonal through that corner.

By the square's symmetry, every corner charge experiences an identical situation, so it suffices to analyse one corner. Each corner charge feels a repulsive force from its two adjacent (SIDE) neighbours, each at distance $a=0.03$ m, and one repulsive force from its DIAGONAL neighbour, at distance $a\sqrt2=0.0424$ m. With $q=6\times10^{-8}$ C, $kq^2=9\times10^9\times(6\times10^{-8})^2=9\times10^9\times3.6\times10^{-15}=3.24\times10^{-5}$.

Each SIDE force has magnitude $F_{side}=\dfrac{kq^2}{a^2}=\dfrac{3.24\times10^{-5}}{(0.03)^2}=\dfrac{3.24\times10^{-5}}{9\times10^{-4}}=0.036$ N. The two side forces act along the two edges meeting at the corner, i.e. at $90^\circ$ to each other, so their resultant has magnitude $F_{side}\sqrt2=0.036\times1.414\approx0.0509$ N, directed exactly along the square's DIAGONAL through that corner (pointing away from the centre, since all charges repel).

The DIAGONAL force has magnitude $F_{diag}=\dfrac{kq^2}{(a\sqrt2)^2}=\dfrac{kq^2}{2a^2}=\dfrac{3.24\times10^{-5}}{2(9\times10^{-4})}=\dfrac{3.24\times10^{-5}}{1.8\times10^{-3}}=0.018$ N, also directed exactly along the same diagonal (away from the centre).

Because the resultant of the two side forces and the diagonal force both point along the SAME diagonal line, they simply ADD directly: $F_{total}=0.0509+0.018=0.0689\,\text{N}=6.89\times10^{-2}\,\text{N},$ matching the source's printed answer exactly, directed radially outward along the diagonal through that corner charge. By the symmetry of the square, every one of the four corner charges experiences this same magnitude, each pointing outward along its own diagonal. [!ANSWER] Resultant force on each corner charge $\approx6.89\times10^{-2}$ N, directed radially outward along the square's diagonal through that corner.

Split each corner charge's net force into the vector sum of two equal perpendicular side-neighbour forces (combine via $\sqrt2$) plus one diagonal-neighbour force (distance $a\sqrt2$), noting all three act along the same diagonal line.

Forgetting the diagonal distance is $a\sqrt2$ (using $2a$ instead), or adding all three forces as simple scalars without recognising the two side forces are perpendicular and must be combined via $\sqrt2$ before adding the diagonal contribution.