A 6 mC charge experiences a force of 6 N directed north. What is the electric field (magnitude and direction) at the location of the charge?
Would E=1000 N/C and direction is north
Find the force (magnitude and direction) on a -20 mC charge that replaces the 6 mC charge.
And F=20 N and direction is south?
To find the electric field at the location of the charge, you can use the formula:
E = F / q
where E is the electric field, F is the force experienced by the charge, and q is the magnitude of the charge. In this case, the force is 6 N and the charge is 6 mC, or 6 × 10^-3 C.
Substituting these values into the formula, we get:
E = 6 N / (6 × 10^-3 C)
E = 1000 N/C
Therefore, the magnitude of the electric field is 1000 N/C.
Since the force experienced by the 6 mC charge is directed north, the electric field at the location of the charge will also be directed north.
Now, to find the force on a -20 mC charge that replaces the 6 mC charge, we can use the formula:
F = Eq
where F is the force on the charge, E is the electric field, and q is the magnitude of the charge. In this case, the electric field is 1000 N/C and the charge is -20 mC, or -20 × 10^-3 C.
Substituting these values into the formula, we get:
F = (1000 N/C) × (-20 × 10^-3 C)
F = -20 N
Therefore, the magnitude of the force on the -20 mC charge is 20 N.
Since the force experienced by the 6 mC charge is directed north, the force on the -20 mC charge will be directed in the opposite direction, which is south.
So, the force on the -20 mC charge is 20 N directed south.