Question

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, and q2 = -4.0 µc, x2 = -2.0 cm, y2 = 1.5 cm. (a) find the magnitude and direction of the electrostatic force on q2

Answer 1

To find the magnitude and direction of the electrostatic force on q2, we can use Coulomb's Law. The magnitude of the electrostatic force is approximately 4.83 x 10^-2 N, and its direction is negative.

Step-by-step explanation:

To find the magnitude and direction of the electrostatic force on q2, we can use 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 the magnitude of the electrostatic force is:

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

Where F is the force, k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.

In this case, q1 = +3.3 µC, q2 = -4.0 µC, and the distance between them can be found using the Pythagorean theorem.

r^2 = (x2 - x1)^2 + (y2 - y1)^2

Substituting the values, we have:

r^2 = (-2.0 cm - 3.5 cm)^2 + (1.5 cm - 0.5 cm)^2

r^2 = (-5.5 cm)^2 + (1.0 cm)^2

r^2 = 30.25 cm^2 + 1.0 cm^2

r^2 = 31.25 cm^2

r = √(31.25 cm^2)

r ≈ 5.59 cm

Now we can calculate the electrostatic force:

F = (8.99 x 10^9 Nm^2/C^2) * ((3.3 µC) * (-4.0 µC)) / (5.59 cm)^2

F ≈ -4.83 x 10^-2 N

The magnitude of the electrostatic force on q2 is approximately 4.83 x 10^-2 N, and its direction is negative.

Answer 2

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is

`d= √((x_1-x_2)^2+(y_1-y_2)^2)`
substituting the coordinates of the two charges, we get

`d= √((3.5+2)^2+(0.5-1.5)^2)=5.6~cm=0.056~m`

2) Then, we can calculate the electrostatic force between the two charges
`q_1` and
`q_2`, which is given by

`F=k_e (q_1 q_2)/(d^2)`
where
`k_e=8.99\cdot10^(9) Nm^2C^(-2)` is the Coulomb's constant.
Substituting numbers, we get

`F=8.99\cdot10^(9) Nm^2C^(-2) ((3.3\cdot10^(-6)~C) (-4\cdot10^(-6)~C))/((0.056~m)^2) =-37.8~N`
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.