Question

Two charges, +9 µC and +16 µC, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force (in N) on a −7 nC charge when placed at the following locations.

(a) halfway between the two
(b) half a meter to the left of the +9 µC charge
(c) half a meter above the +16 µC charge in a direction perpendicular to the line joining the two fixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the +x-axis.)

Answer 1

To find the magnitude and direction of the net force on a -7 nC charge, we need to calculate the forces exerted by each of the fixed charges on it and then find their vector sum at different locations.

Step-by-step explanation:

To find the magnitude and direction of the net force on a -7 nC charge, we need to calculate the forces exerted by each of the fixed charges on it and then find their vector sum.

(a) Halfway between the two charges:

The magnitude of the force resulting from Coulomb's Law is:

F = k(Q1 * Q2) / r²

Where Q1 = -7 nC, Q2 = 9 µC, k = Coulomb's constant, and r = distance between the charges

The direction of the net force can be found by considering the direction of each individual force and their magnitudes.

(b) Half a meter to the left of the +9 µC charge:

The force from the +9 µC charge will be attractive since the charges have opposite signs. The magnitude and direction can be calculated using Coulomb's Law.

(c) Half a meter above the +16 µC charge in a direction perpendicular to the line joining the two fixed charges:

The force from the +16 µC charge will be repulsive since the charges have the same sign. The magnitude and direction can be calculated using Coulomb's Law.

Answer 2

A) 1.76U×10⁻³N

B) 2.716×10⁻³N

C) 264.5⁰

Step-by-step explanation:

See detailed workings for (a), (b), (c) attached.