Question

Three point charges are arranged in a horizontal line as shown below. Find the electric forces (in units of kQ^2/R^2) on the charges given the following: Q_1 = -80 Q, Q_2 = -4 Q, Q_3 = 36 Q, r_1 = 4 R and r_2 = 2 R Remember that a positive force points to the right and a negative force points to the left. What is the net force on charge Q_1? What is the net force on charge Q_2? The total force on Q_2 is going to be a vector sum of F_2: (the force on Q_2 due to Q_1) and F_23 (the force on Q_2 due to Q_3). You need to find both the magnitudes and directions of F_21 and F_23. The magnitudes of the forces are given by: F_21 = kQ_2Q_1/r_1^2 and F_23 = kQ_2Q_3/r_2^2, where k = 9e9 Vm/C and the absolute values for all charges should be used. The direction of each force is determined by the two charges, where opposite charges attract and same charges repel. What is the net force on charge Q_3? What is the sum of the forces on all three charges?

solve all parts

Answer 1

On charge Q₁ :

F₃ = force by charge Q₃ = k Q₁ Q₃/r₁² = k (80 Q) (36Q)/(4R)² = 180 kQ/R²    towards right

F₂ = force by charge Q₂ = k Q₁ Q₂/(r₁ + r₂ )² = k (80 Q) (4Q)/(6R)² = 8.88 kQ/R²     towards left

Net force on Q₁ = F₃ - F₂ = 180 kQ/R² - 8.88 kQ/R² = 171.12 kQ/R²

On charge Q₂ :

F₃ = force by charge Q₃ = k Q₂ Q₃/r₂² = k (4 Q) (36Q)/(2R)² = 36 kQ/R²    towards left

F₁ = force by charge Q₁ = k Q₁ Q₂/(r₁ + r₂ )² = k (80 Q) (4Q)/(6R)² = 8.88 kQ/R²     towards left

Net force on Q₁ = F₃ + F₁ = 36 kQ/R² + 8.88 kQ/R² = 44.88 kQ/R²    towards left

On charge Q₃ :

F₂ = force by charge Q₂ = k Q₂ Q₃/r₂² = k (4 Q) (36Q)/(2R)² = 36 kQ/R²    towards right

F₁ = force by charge Q₁ = k Q₁ Q₃/(r₁ )² = k (80 Q) (36Q)/(4R)² = 180 kQ/R²     towards left

Net force on Q₃ = F₁ - F₂ = 180 kQ/R² - 36 kQ/R² = 144 kQ/R²    towards left