A test charge of +3.00 10-6 C experiences a downward electrostatic force of 15 N when placed at a certain point in space. What are the magnitude and direction of the electric field at this point?
force=Eq
Direction of E is in the direction a positive test charge is deflected, in this case, "downward".
To find the magnitude and direction of the electric field at a certain point due to a charge, you can use Coulomb's Law.
Coulomb's Law states that the force between two charges is given by the formula:
F = k * (|q1| * |q2|) / r^2
Where:
- F is the force between the charges,
- k is the electrostatic constant (k ≈ 9.0 x 10^9 N m^2/C^2),
- |q1| and |q2| are the magnitudes of the charges,
- r is the distance between the charges.
In this case, we have a test charge of +3.00 × 10^-6 C experiencing a force of 15 N. We need to find the electric field at this point.
The electric field (E) at a point is defined as the force per unit charge. Mathematically, we can express it as:
E = F / q
Where:
- E is the electric field,
- F is the force on the test charge at that point,
- q is the magnitude of the test charge.
In this case, we are given the force and the charge. So, we can substitute these values into the equation to calculate the electric field.
E = F / q
= 15 N / 3.00 × 10^-6 C
= 5.00 × 10^6 N/C
So, the magnitude of the electric field at this point is 5.00 × 10^6 N/C. Since the force is acting downward on a positive charge, the direction of the electric field is also downward.