Question

Two point charges are fixed on the y axis: a negative point charge q1 = -24 µC at y1 = +0.19 m and a positive point charge q2 at y2 = +0.33 m. A third point charge q = +8.0 µC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 18 N and points in the +y direction. Determine the magnitude of q2.

Answer 1

To determine the magnitude of q2, we can use Coulomb's Law to calculate the net force exerted on charge q. By setting up an equation using the given information, we can find that the magnitude of q2 is approximately 4.84 μC.

Step-by-step explanation:

To determine the magnitude of q2, we can use Coulomb's Law, which states that the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, we have two charges, q1 and q2, with a distance of 0.33 - 0.19 = 0.14m between them.

The net electrostatic force on charge q can be calculated using the equation F = k * (|q1 * q| / r1^2) + k * (|q2 * q| / r2^2), where k is the electrostatic constant (8.99 x 10^9 N*m^2/C^2).

From the given information, we know that F = 18 N and points in the +y direction, so we can set up the following equation: 18 = k * (|(-24 μC) * (8 μC)| / (0.19m)^2) + k * (|q2 * (8 μC)| / (0.33m)^2).

Solving this equation, we can find the magnitude of q2, which is approximately 4.84 μC.

Answer 2

4.51 * 10^{-5} C

Step-by-step explanation:

The force between two charge q and q1 is given as

`F = (k*q*q1)/(r^2)`

`= ((9.0 * 10^9)(24 * 10^(-6))(8 * 10^(-6) C))/((0.19m)^2)`

= 47.86 N in the +y direction

We need the force between q and q2 to be (47.86 - 18) = 29.86 N in the other direction to get the desired result.

solving for q2,

`q2 = (Fr^2)/((kq))`

`= ((29.86)(0.33 m)^2)/((9.0 * 10^9*8*10^(-6) C))`

`= 4.51 * 10^(-5) C`