Question

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm , y = 3.2 nm . Part A Find the force on the ion of charge -2e. Express your answer in vector components, separated by commas, using two significant figures.

Answer 1

The electric force on the -2e ion caused by the -3e ion can be calculated using Coulomb's Law, considering the direction and magnitude of the force and accounting for the repulsive nature of the interaction between two like-charged ions.

Step-by-step explanation:

The student's question focuses on calculating the electric force on a charge due to the presence of a nearby charged ion, using Coulomb's Law. The given scenario involves two negative ions with charges of -2e and -3e. The force experienced by the ion at the origin (charge -2e) due to the second ion (charge -3e), located at (3.8 nm, 3.2 nm), can be calculated by determining the electric force vector using the equation F = k * (q1 * q2) / r^2, where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.

To find the vector components, we first calculate the magnitude of the force and then determine the direction using the coordinates given. Since the charges are both negative, the force between them will be repulsive, which means the -2e charge will experience a force in the direction away from the -3e charge. The force components are found by projecting this force onto the x and y axes.

Answer 2

`F = 5.6 * 10^(-11) N`

`F_x = 4.28* 10^(-11) N`

`F_y = 3.61* 10^(-11) N`

Step-by-step explanation:

Magnitude of the force on -2e charge due to another charge -3e is given as

`F = (kq_1q_2)/(r^2)`

here we will have

`F = ((9* 10^9)(2e)(3e))/((3.8^2 + 3.2^2)* 10^(-18))`

`F = 5.6 * 10^(-11) N`

now for the two components of the force we know that this force always along the line joining two charges

so we will have

`tan\theta = (y)/(x)`

`\theta = tan^(-1)(3.2 nm)/(3.8 nm)`

`\theta = 40.1 degree`

now the two components of above force will be

`F_x = Fcos\theta`

`F_x = (5.6* 10^(-11))cos40.1`

`F_x = 4.28* 10^(-11) N`

`F_y = Fsin\theta`

`F_y = (5.6* 10^(-11))sin40.1`

`F_y = 3.61* 10^(-11) N`