Question

Calculate the resultant force at Q3
IF
Q1=-16 μC
distances from Q1 TO Q2 is =0.6m
Q2=6 μC
distances from Q1 to Q2 is =0.3m
Q3=-8 μC
IN SIMPLE STEPS

Answer 1

To calculate the resultant force at Q3, we need to consider the forces from Q1 and Q2 on Q3. The force between two charges can be calculated using Coulomb's law:

F = k * (Q1 * Q2) / d^2

where F is the force, k is the electrostatic constant (9 x 10^9 N*m^2/C^2), Q1 and Q2 are the charges, and d is the distance between the charges.

First, let's calculate the force from Q1 on Q3:
F1 = k * (Q1 * Q3) / d1^2

Given Q1 = -16 μC, Q3 = -8 μC, and d1 = 0.6m:

F1 = (9 x 10^9 N*m^2/C^2) * (-16 μC * -8 μC) / (0.6m)^2

F1 = (9 x 10^9 N*m^2/C^2) * (128 x 10^-12 C^2) / (0.36m^2)

F1 = (9 x 10^9 N*m^2/C^2) * (1.28 x 10^-10) / 0.1296

F1 = 88.8889 N

Next, let's calculate the force from Q2 on Q3:
F2 = k * (Q2 * Q3) / d2^2

Given Q2 = 6 μC, Q3 = -8 μC, and d2 = 0.3m:

F2 = (9 x 10^9 N*m^2/C^2) * (6 μC * -8 μC) / (0.3m)^2

F2 = (9 x 10^9 N*m^2/C^2) * (-48 x 10^-12 C^2) / (0.09m^2)

F2 = (9 x 10^9 N*m^2/C^2) * (-4.8 x 10^-10) / 0.0081

F2 = -400000 N

Now, to find the resultant force at Q3, we need to add the forces F1 and F2:

Resultant force (FR) = F1 + F2

FR = 88.8889 N + (-400000 N)

FR = -399911.1111 N

Therefore, the resultant force at Q3 is approximately -399911.1111 N.