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?
I know for the first question it's 7500 north but I don't understand the second portion. How do you calculate the magnitude and direction if the charge is removed?
When the charge was removed, the electric field became zero.
To calculate the electric field magnitude and direction at the location where the charge used to be, we need to consider the concept of electric field due to a single point charge.
The electric field created by a point charge at a certain location is defined as the force experienced by a positive test charge placed at that location, divided by the magnitude of the test charge. Mathematically, the electric field (\(E\)) due to a point charge (\(Q\)) at a distance (\(r\)) from the charge is given by the equation:
\[ E = \dfrac{K \cdot Q} {r^2} \]
where \(K\) is the electrostatic constant, approximately equal to \(9 \times 10^9\) Nm\(^2\)/C\(^2\).
So, to calculate the electric field at the location where the charge used to be, we need to consider the charge that created the field, its distance from the location, and the magnitude of the charge.
However, when the charge is removed, the electric field at that location becomes zero. This is because the electric field is a property of the charge configuration itself, and when the charge is removed, there is no charge to create an electric field. Therefore, the electric field at the location it used to occupy becomes zero in magnitude and direction.