Question

a 15-nc point charge is placed on the x axis at x = 1.5 m, and a -20-nc charge is placed on the y axis at y = -2.0m. what is the magnitude of the electric field at the origin

Answer 1

The magnitude of the electric field at the origin is 135 N/C.

Step-by-step explanation:

To find the magnitude of the electric field at the origin, we need to calculate the electric field contribution from each charge and then combine them.

Let's start by calculating the electric field due to the +15 nC charge at x = 1.5m. The formula for the electric field from a point charge is given by:

E = k * q / r^2

Where E is the electric field, k is the electrostatic constant (9 x 10^9 N m^2 / C^2), q is the charge, and r is the distance from the charge.

Plugging in the values, we have:

E1 = (9 x 10^9 N m^2 / C^2) * (15 x 10^-9 C) / (1.5m)^2

Simplifying the expression, we get:

E1 = 90 N/C

Next, let's calculate the electric field due to the -20 nC charge at y = -2.0m. Since the charge is on the y-axis, the distance from the charge to the origin is 2.0m. Using the same formula, but with a negative charge, we have:

E2 = (9 x 10^9 N m^2 / C^2) * (-20 x 10^-9 C) / (2.0m)^2

Simplifying the expression, we get:

E2 = -225 N/C

Finally, to find the net electric field at the origin, we add the two electric field vectors:

E_total = E1 + E2

Plugging in the values, we have:

E_total = 90 N/C + (-225 N/C)

Simplifying the expression, we get:

E_total = -135 N/C

Therefore, the magnitude of the electric field at the origin is 135 N/C.