HOW TO SOLVE? Net Force (Physics)

Question

Three identical point charges , each 10 mC , are placed at the vertices of an equilateral triangle such that

two of these charges are located at the coordinate origin and at (10 cm, 0 cm) and the other charge

is located at point P whose coordinates are both positive. Determine the net force on charge placed at

point P.

Answer 1

Identical charges are going to repel each other, hence the net force on the charge placed at point P will be directed upwards, away from the X axis, where the other two charges are located.

Let us denote the force with which the charge at origin repel the charge at point P as F. It’s magnitude can easily be calculated from the Coulomb’s law.

F = k q² / r², where k is the Coulomb’s constant 8.988×10⁹ N⋅m²⋅C⁻², q = 0.01 C, and r = 0.1 m

Substituting the above values into the equation we obtain

F = 8.988×10⁷ N

The direction of the force is 60°, which means that its horizontal component F_x = F cos 60° = F/2, and its vertical component F_y = F sin 60° = F √3 /2

Similarly the force with which the charge at (10,0) repels the charge at point P has the same magnitude F, but the components will be F_x = - F cos 60° = - F/2, and F_y = F sin 60° = F √3 /2

Hence when the two force are added together, their horizontal components will cancel each other out. The resulting net force will thus be pointing upwards, and its magnitude will be F √3 = 1.557×10⁸ N