a positive test charge of 3.0uC experiences a force of 0.75 N in an electric field. What is the magnitude of the electric field at the location of the test charge?

To find the magnitude of the electric field at the location of the test charge, you can use the equation:

Electric field strength (E) = Force (F) / Test charge (q)

In this case, the force experienced by the test charge (F) is 0.75 N, and the test charge (q) is 3.0 μC (microcoulombs or x 10^-6 C).

First, convert the test charge from microcoulombs to coulombs:

3.0 μC = 3.0 x 10^-6 C

Now, plug the values into the equation:

E = 0.75 N / (3.0 x 10^-6 C)

E = 0.75 N / 3.0 x 10^-6 C

E ≈ 2.5 x 10^5 N/C

Therefore, the magnitude of the electric field at the location of the test charge is approximately 2.5 x 10^5 N/C.

10