A -10 mC charge experiences a force of 75 N directed south. What is the electric field magnitude and direction at the location of the charge?
If this charge is removed and not replaced, what is the electric field magnitude and direction at the location it used to occupy?
To determine the electric field magnitude and direction at the location of the charge, we can use Coulomb's law, which states that the force between two charged objects is given by the equation:
F = k * (|q1| * |q2|) / r^2
where F is the force, k is the electrostatic constant (k ≈ 9.0 x 10^9 N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between them.
In this case, the charge experiencing the force is -10 mC (negative indicating the charge is negative), and the force is 75 N directed south. Let's assume that the force is due to an unknown positive charge, q2.
First, we need to find the magnitude of q2. Rearranging Coulomb's law, we have:
|q2| = (F * r^2) / (k * |q1|)
Plugging in the given values, we have:
|q2| = (75 N * (1 meter)^2) / (9.0 x 10^9 N m^2/C^2 * 10^(-5) C)
Calculating this expression, we find that |q2| ≈ 8.3 x 10^(-4) C.
Next, to find the electric field at the location of the charge, we can use the equation:
E = F / |q1|
where E is the electric field.
Substituting the values, we have:
E = 75 N / (10^(-5) C)
Calculating this expression, we find that E = 7.5 x 10^6 N/C.
Therefore, the electric field magnitude at the location of the charge is 7.5 x 10^6 N/C, directed south.
Now, if the charge is removed and not replaced, the electric field at the location it used to occupy will be zero. This is because the electric field at a point in space due to a charged object depends on the presence and magnitude of the charge. If the charge is removed, there is no source of the electric field, resulting in a net electric field of zero at that location.