The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = +8.70 µC; the other two charges have identical magnitudes, but opposite signs: q2 = -4.50 µC and q3 = +4.50 µC.
Determine the net force (magnitude and direction) exerted on q1 by the other two charges.
If q1 had a mass of 1.62 g and it were free to move, what would be its acceleration?
To find the net force exerted on q1 by the other two charges, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. It can be represented as:
F = k * (|q1| * |q2|) / r^2
Where F is the force, k is Coulomb's constant (8.99 × 10^9 N·m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between them.
In this case, we have two charges interacting with q1:
For q1 and q2:
F1 = k * (|q1| * |q2|) / r1^2
For q1 and q3:
F2 = k * (|q1| * |q3|) / r2^2
Since q2 and q3 have opposite signs, the forces will have opposite directions. Therefore, we need to find the vector sum of F1 and F2 to determine the net force on q1.
To find the magnitude of the net force, we can use the Pythagorean theorem:
Net force magnitude = sqrt(F1^2 + F2^2)
To find the direction of the net force, we can use trigonometry. The angle θ between the net force and the positive x-axis can be found using:
θ = tan^(-1)(F2/F1)
Now, to calculate the acceleration of q1 if it were free to move, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
F = m * a
In this case, q1 has a charge and not a mass. However, we can substitute the magnitude of the electric force (|F|) calculated earlier for F in the equation above to find the acceleration.
a = |F| / m
Plugging in the values, we can now determine the answers.