Thirty microcoulombs of NEGATIVE charge experiences an electrostatic force of 27. mN. What is the magnitude and direction of the electric field?
To determine the magnitude and direction of the electric field, we can use Coulomb's Law and the fact that force equals the product of charge and electric field.
Coulomb's Law states that the electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them:
F = k * (q1 * q2) / r^2
Where:
F is the electrostatic force,
k is the electrostatic constant (k ≈ 8.99 x 10^9 N m^2/C^2),
q1 and q2 are the charges of the objects, and
r is the distance between the charges.
In this case, we have q1 = 30 μC (negative charge) and q2 = unknown (electric field). The force F is given as 27. mN (millinewtons), which we need to convert to Newtons (N):
27. mN = 27. x 10^-3 N
Now, we can rearrange Coulomb's Law to solve for the electric field:
F = q1 * E => E = F / q1
E = (27. x 10^-3 N) / (30 x 10^-6 C)
E = 0.9 N/C
The magnitude of the electric field is 0.9 N/C.
To determine the direction of the electric field, we need to know the direction of the force. Since the negative charge experiences a force, we know that the electric field points away from the charge, opposite to the direction of the force.
Therefore, the direction of the electric field is opposite to the force, pointing towards the positive charge (assuming the force is attractive due to opposite charges).
So, the magnitude of the electric field is 0.9 N/C, and its direction is opposite to the force, which is towards the positive charge.