Question

Two point charges are located on the y axis as follows: charge q1 = -2.30 nC at y1 = -0.600 m , and charge q2 = 2.80 nC at the origin (y = 0). What is the magnitude of the net force exerted by these two charges on a third charge q3 = 7.50 nC located at y3 = -0.300 m ?

Answer 1

The magnitude of the net force exerted by the two charges on the third charge is 1.38 N.

Step-by-step explanation:

The magnitude of the net force exerted by q1 and q2 on q3 can be found using Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k(q1 * q2) / r^2

Where F is the force, k is Coulomb's constant (k = 9 x 10^9 N * m^2 / C^2), q1 and q2 are the charges, and r is the distance between them.

In this case, q1 = -2.30 nC, q2 = 2.80 nC, q3 = 7.50 nC, y1 = -0.600 m, y3 = -0.300 m, and y2 = 0 m.

Using the formula and substituting the values, we can calculate the force:

F = (9 x 10^9 N * m^2 / C^2) * (q1 * q3) / ((y3 - y1)^2)

F = (9 x 10^9 N * m^2 / C^2) * (-2.30 nC * 7.50 nC) / ((-0.300 m + 0.600 m)^2)

F = -1.38 N

Therefore, the magnitude of the net force exerted by the two charges on the third charge is 1.38 N.

Answer 2

3.825*10^-6 N

Step-by-step explanation:

As particle 1 and particle 3 has opposite types of charge, particle 3 will be attracted to particle 1. And as particle 2 and 3 has the same sign, they will repel each other. Due to the position of the particles, both the Force that 1 exerts on 3 and the force that 2 exerts on 3, will have the same direction. Now, you need the magnitude. You can use the following expression:

`F_e = K(q_1*q_3)/(r^2)`

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and q2 is the charge of the particles, and r is the distance.

Force 1 on 3 is equal to:

`F_e = K(q_1*q_3)/(r^2) = 9*10^9 (Nm^2)/(C^2) * (2.3*10^(-9)C * 7.5*10^(-9)C)/((-0.6m - (-0.3m))^2) = 1.725 * 10^(-6) N`

Force 2 on 3:

`F_e = K(q_2*q_3)/(r^2) = 9*10^9 (Nm^2)/(C^2) * (2.8*10^(-9)C * 7.5*10^(-9)C)/((0m - (-0.3m))^2) = 2.1 * 10^(-6) N`

The magnitude of the resultant force is the addition of both forces:

`F_e = 1.725*10^(-6) N + 2.1*10^(-6) N= 3.825 * 10^(-6) N`