three point charges lie on the same x-axis. charge 1 (-2.1 μC) is at the origin, charge 2 (+3.2 μC) is at x = 7.5 cm, and charge 3 (-1.8 μC) is at x = 11 cm. what are the direction and the magnitude of the total force exerted on charge 1.
Answers
Given info : Three point charges lie on the same x-axis. charge 1 (-2.1 μC) is at the origin, charge 2 (+3.2 μC) is at x = 7.5 cm, and charge 3 (-1.8 μC) is at x = 11 cm.
To find : the direction and magnitude of the total force exerted on the charge 1 are..
solution :
O(-2.1μC)--------A(+3.2μC) ------B(-1.8μC)
O Is origin, A = (7.5, 0) and B = (11, 0)
so, OA = 7.5 cm , OB = 11 cm
force due to charge 2 on charge 1 , F₂₁ = K(2.1μC)(3.2μC)/(7.5cm)² [ along OA ]
= (9 × 10^9 × 2.1 × 3.2 × 10^-12)/(56.25 × 10^-4)
= 10.752 N
in vector form, F₂₁ = 10.752 i
force due to charge 3 on charge 1, F₃₁ = K(1.8μC)(2.1μC)/(11cm)² [ along BO ]
= (9 × 10^9 × 1.8 × 2.1 × 10^-12)/(121 × 10^-4)
= 2.811 N
in vector form, F₃₁ = -2.811 i
net force, F = F₂₁ + F₃₁
= 10.752 i - 2.811 i
= 7.941 i ≈ 8 i
Therefore net force acting on the charge 1 is 8 N (approx) and direction of force is along positive x-axis.