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 the direction still be north? And the magnitude 1000?

E = F/q = 6/6•10^-3 =100 V/m.

The direction still be north.
magnitude of the electric field is
E = k •q /r²,
therefore, E ~ 1/r²

To calculate the electric field at the location of the charge, you can use the formula:

Electric Field (E) = Force (F) / Charge (q)

Given that the charge is 6 mC and experiences a force of 6 N, we can plug these values into the equation to find the electric field.

E = 6 N / 6 mC

First, let's convert the charge from milliCoulombs (mC) to Coulombs (C):

1 C = 1000 mC

Therefore, 6 mC = 6/1000 C = 0.006 C.

Now, we can substitute the values into the equation:

E = 6 N / 0.006 C

E = 1000 N/C

The magnitude of the electric field is indeed 1000 N/C.

Regarding the direction of the electric field, it is not necessarily in the same direction as the force experienced by the charge. The electric field direction is determined by the positive charge convention. Conventionally, the direction of the electric field is taken to be the direction a positive charge would move if placed in that field. In this case, since the charge experiences a force directed north, it means the electric field is directed towards the south. Therefore, the direction of the electric field is south.