Two point charges are located as shown in the figure, with charge q1 = +3.2 C at x = − 3.0 m, y = 0, and charge q2 = +2.5 C at x = +1.0 m, y = +2.0 m. An electron is now taken from a point very far away and placed at the origin. How much work must be done on the electron to move it to the origin?

1 J

Please show calculations, I am at a lost as how to begin...Thank you

Why don't you find the potential at x=0?

Then add the potential contributions from each of the charge.

Then work= potential*e

6. Two charges are placed as shown in figure. The magnitude of q1 is 3.00 µC, but its sign and the value of the charge q2 are not known. The

direction of the net electric field E at point P is entirely in the negative y-direction. (a) Considering the different possible signs of q1 and q2 , there are four possible diagrams that could represent the electric fields E1 and E2 produced by q1 and q2 . Sketch the four possible electric-field configurations. (b) Using the sketches from part (a) and the direction of E , deduce the signs of q1 and q2. (c) Determine the magnitude of E .

To find the work done on the electron to move it to the origin, we need to calculate the electric potential energy of the electron at the initial position and at the final position. The work done on the electron is the change in electric potential energy.

The potential energy of a point charge in an electric field is given by the equation:

PE = k * (|q1| * |q2|) / r

where k is the electrostatic constant (9 × 10^9 Nm^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.

At the initial position, the distance between the electron and charge q1 is given by:

r1 = sqrt((x1 - x2)^2 + (y1 - y2)^2)

where (x1, y1) are the coordinates of the electron (0, 0), and (x2, y2) are the coordinates of charge q1 (-3.0, 0).

Substituting the given values:

r1 = sqrt((0 - (-3.0))^2 + (0 - 0)^2) = sqrt(9 + 0) = 3.0 m

At the final position (origin), the distance between the electron and charge q2 is given by:

r2 = sqrt((x1 - x3)^2 + (y1 - y3)^2)

where (x3, y3) are the coordinates of charge q2 (1.0, 2.0).

Substituting the given values:

r2 = sqrt((0 - 1.0)^2 + (0 - 2.0)^2) = sqrt(1 + 4) = sqrt(5) m

Now we can calculate the potential energy at both positions:

PE1 = k * (|q1| * |q2|) / r1
PE1 = (9 × 10^9 Nm^2/C^2) * (3.2 C * 2.5 C) / 3.0 m

PE2 = k * (|q1| * |q2|) / r2
PE2 = (9 × 10^9 Nm^2/C^2) * (3.2 C * 2.5 C) / sqrt(5) m

The work done on the electron is the difference between the potential energies:

Work = PE2 - PE1

Calculating the work:

Work = [(9 × 10^9 Nm^2/C^2) * (3.2 C * 2.5 C) / sqrt(5) m] - [(9 × 10^9 Nm^2/C^2) * (3.2 C * 2.5 C) / 3.0 m]

Simplifying the expression:

Work = [(9 × 10^9 Nm^2/C^2) * (3.2 C * 2.5 C) / sqrt(5) m] - [(9 × 10^9 Nm^2/C^2) * (3.2 C * 2.5 C) / 3.0 m]

Work = (22.8 × 10^9 Nm) / sqrt(5) m - (22.8 × 10^9 Nm) / 3.0 m

Calculating this further:

Work = (22.8 × 10^9 Nm) / sqrt(5) m - (22.8 × 10^9 Nm) / 3.0 m

Work = (150.164 × 10^9 Nm) / sqrt(5) m - (7.6 × 10^9 Nm) / m

Finally, we can simplify and calculate the work:

Work = (150.164 × 10^9 Nm - 7.6 × 10^9 Nm) / sqrt(5) m

Work = (142.564 × 10^9 Nm) / sqrt(5) m

Work ≈ 142.564 × 10^9 / sqrt(5) J

Calculating this:

Work ≈ 142.564 / 2.236 J

Work ≈ 63.8 J (rounded to one decimal place)

Therefore, the work done on the electron to move it to the origin is approximately 63.8 J.