What is the magnitude and direction of an electric field that exerts a 4.00 ✕ 10 -5 N upward force on a - 1.80 µC charge?
N/C
To find the magnitude and direction of an electric field, we can use Coulomb's Law and the equation for electric field.
Coulomb's Law states that the force between two charged objects is given by the equation:
F = k * (q1 * q2) / r^2
Where:
- F is the force between the charges (in Newtons),
- k is Coulomb's constant (approximately 9 * 10^9 N·m^2/C^2),
- q1 and q2 are the charges on the objects (in Coulombs), and
- r is the distance between the charges (in meters).
In this case, we are given the force and one of the charges, and we need to find the electric field. The equation for electric field is given by:
E = F / q
Where:
- E is the electric field (in N/C),
- F is the force (in Newtons), and
- q is the charge (in Coulombs).
First, we can rearrange Coulomb's Law equation to solve for the electric field:
F = k * (q1 * q2) / r^2
F = E * q
E = F / q
Using these equations, we can find the electric field:
E = (4.00 * 10^-5 N) / (-1.80 * 10^-6 C)
Calculating this gives:
E = -2.22 * 10^4 N/C
The negative sign indicates that the electric field is pointing in the opposite direction of the force, which means it is downward in this case.
Therefore, the magnitude of the electric field is 2.22 * 10^4 N/C, and the direction is downward.