A charge of +6.0µC experiences a force of 2.0 mN in the +x-direction at a certain point in space. (a) What was the electric field there before the charge was placed there? (b) Describe the force a - 2.0µC charge would experience if it were used in place of the +6.0µC charge.
To find the electric field at a certain point in space due to a charge, you can use Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
(a) To find the electric field before the charge was placed there, we can rearrange Coulomb's law equation to solve for the electric field. The equation for the electric field (E) is given by:
E = F / q
Where:
E is the electric field.
F is the electric force.
q is the charge experiencing the force.
In this case,
F = 2.0 mN (2.0 x 10^-3 N)
q = +6.0 µC (6.0 x 10^-6 C)
Plugging these values into the equation, we get:
E = (2.0 x 10^-3 N) / (6.0 x 10^-6 C)
E ≈ 333.33 N/C
So, the electric field at that point before the charge was placed there was approximately 333.33 N/C in the +x-direction.
(b) To determine the force a -2.0µC charge would experience if it were used in place of the +6.0µC charge, we can use the same equation and the same electric field value we found earlier.
F = Eq
q = -2.0 µC (-2.0 x 10^-6 C)
Plugging these values into the equation, we get:
F = (333.33 N/C) * (-2.0 x 10^-6 C)
F ≈ -0.67 mN
So, a -2.0µC charge would experience a force of approximately -0.67 mN in the +x-direction. The negative sign indicates that the force would be in the opposite direction of the electric field.