23.1 .. A point charge q 1 = +2.40 mC is held stationary at the origin. A second point charge q 2 = -4.30 mC moves from the point x = 0.150 m, y = 0 to the point x = 0.250 m, y = 0.250 m. How much work is done by the electric force on q 2 ?

To find the work done by the electric force on q2, we can use the formula:

Work = Force * distance

First, we need to find the force exerted by the electric field. The force between two charges can be calculated using Coulomb's Law:

F = k * q1 * q2 / r^2

where F is the force, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.

In this case, q1 = +2.40 mC and q2 = -4.30 mC.

Next, we need to find the distance covered by q2. The distance can be calculated using the Pythagorean theorem:

Distance = √((x2 - x1)^2 + (y2 - y1)^2)

In this case, (x1, y1) = (0.150 m, 0) and (x2, y2) = (0.250 m, 0.250 m).

Now, let's calculate the force and distance:

Distance = √((0.250 - 0.150)^2 + (0.250 - 0)^2) = √((0.1)^2 + (0.25)^2) = √(0.01 + 0.0625) = √0.0725 = 0.269 m

Force = (8.99 x 10^9 N m^2/C^2) * (+2.40 x 10^-3 C) * (-4.30 x 10^-3 C) / (0.269 m)^2

Force = -3.23 N (negative sign indicates an attractive force)

Now, we can calculate the work done:

Work = Force * distance

Work = -3.23 N * 0.269 m

Work = -0.87087 J

Therefore, the work done by the electric force on q2 is approximately -0.87087 Joules.

To calculate the work done by the electric force on q2, we can use the formula:

Work = Force * Distance * cos(theta)

where Force is the electric force between q1 and q2, Distance is the displacement of q2, and theta is the angle between the force and displacement vectors.

The electric force between two point charges can be calculated using Coulomb's Law:

Electric Force = (k * |q1 * q2|) / r^2

where k is Coulomb's constant, q1 and q2 are the magnitudes of the two charges, and r is the distance between them.

First, let's calculate the electric force between q1 and q2 using Coulomb's Law. Given:
q1 = +2.40 mC = 2.40 * 10^-3 C
q2 = -4.30 mC = -4.30 * 10^-3 C
k = 9.0 * 10^9 N * m^2 / C^2

We can calculate the distance between the two points using the Pythagorean theorem:
Distance = sqrt((x2-x1)^2 + (y2-y1)^2)

Given:
x1 = 0
y1 = 0
x2 = 0.250 m
y2 = 0.250 m

Substituting these values into the equation, we get:
Distance = sqrt((0.250 - 0)^2 + (0.250 - 0)^2)
Distance = sqrt(0.125 + 0.125)
Distance = sqrt(0.250)
Distance = 0.5 m

Now let's calculate the electric force F:
F = (k * |q1 * q2|) / r^2
F = (9.0 * 10^9 * |2.40 * 10^-3 * -4.30 * 10^-3|) / (0.5)^2
F = (9.0 * 10^9 * 2.40 * 10^-3 * 4.30 * 10^-3) / 0.25
F = 97.92 N (approximately)

Next, we need to find the angle theta between the force and displacement vectors. Since the force and displacement vectors are parallel along the x-axis, the angle between them is 0 degrees.

Now we can calculate the work done:
Work = Force * Distance * cos(theta)
Work = 97.92 * 0.5 * cos(0)
Work = 48.96 J (approximately)

Therefore, the work done by the electric force on q2 is approximately 48.96 Joules.

Find the electric potential at both points.

v1=kq1/x=k(2.4e-3)/.150
v2=kq1/(sqrt(.250^2+.250^2)

now the work is (v2-V1)(-4.3e-3), where q1 is 2.40e-3 C, and k is 9.0 x 109