what is the magnited of electric field E at a field point 2.0 m form a point charge q -4.0nc
To find the magnitude of the electric field at a certain point due to a point charge, you can use Coulomb's Law. Coulomb's Law states that the electric field created by a point charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the charge and the point.
The formula for the magnitude of the electric field (E) caused by a point charge (q) at a distance (r) can be expressed as:
E = k * (|q| / r^2)
Where:
- E represents the magnitude of the electric field.
- k is the electrostatic constant, approximately equal to 9.0 × 10^9 N m^2/C^2.
- |q| is the magnitude of the charge.
- r is the distance from the charge to the point in meters.
In your case, the charge (q) is given as -4.0 nC (nanocoulombs), and the distance (r) is 2.0 meters.
First, convert the charge from nanocoulombs to coulombs:
-1 nC = -1 × 10^(-9) C
Therefore, -4.0 nC = -4.0 × 10^(-9) C
Now, substitute the values into the electric field formula:
E = (9.0 × 10^9 N m^2/C^2) * (|-4.0 × 10^(-9) C| / (2.0 m)^2)
Simplifying the equation yields:
E = (9.0 × 10^9 N m^2/C^2) * (4.0 × 10^(-9) C) / (4.0 m^2)
E = 9.0 × 10^9 N m^2 / C^2 * 10^(-9) C / 1 m^2
E = 9.0 N/C
Therefore, the magnitude of the electric field (E) at a point 2.0 meters away from a point charge of -4.0 nC is 9.0 N/C.