A charge, q1 = +4 nC, experiences a force of 3x10-5 N to the east when placed in an electric field. If the charge is replaced by another, q2 = -12 nC, what will be the magnitude and direction of the force on the charge at that position?
To find the magnitude and direction of the force on the charge at the given position, we need to use the formula for the electric force:
F = q * E
where F is the force, q is the charge, and E is the electric field.
Given:
q1 = +4 nC
F1 = 3x10^-5 N
First, let's find the electric field, E, at that position using q1 and F1:
E = F1 / q1
Substituting the values:
E = (3x10^-5 N) / (+4 nC)
Next, we can calculate the force, F2, on the second charge, q2 = -12 nC, by multiplying q2 and E:
F2 = q2 * E
Substituting the values:
F2 = (-12 nC) * [(3x10^-5 N) / (+4 nC)]
Now, let's calculate the magnitude of the force:
|F2| = |(-12 nC) * [(3x10^-5 N) / (+4 nC)]|
|F2| = |-12 nC * (3x10^-5 N / 4 nC)|
|F2| = |-36x10^-5 N / 4|
|F2| = |-9x10^-5 N|
So, the magnitude of the force on the charge at that position is 9x10^-5 N.
Finally, to determine the direction of the force, we look at the sign of the charge q2. Since q2 is negative, the force will act in the opposite direction to the electric field, which is to the east in this case.
Therefore, the magnitude of the force is 9x10^-5 N, and the direction is to the west (opposite to the electric field).