A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south. What is the charge of object A?
Object A must have a negative charge if the direction is south, at a location north of the charge.
The charge Q is given by
E = 40 = -k Q/R^2, where R = 0.25 m
Solve for Q
k is the Coulomb constant.
To determine the charge of object A, we can use Coulomb's law, which states that the electric field created by a point charge is given by the equation:
E = k * (Q / r^2)
Where:
E is the electric field strength,
k is the electrostatic constant (approximately 8.99 x 10^9 N m^2/C^2),
Q is the charge of the object,
and r is the distance between the object and the point.
In this case, we are given the electric field strength (E = 40.0 N/C) and the distance (r = 0.250 m). We can rearrange the equation to solve for the charge (Q):
Q = E * r^2 / k
Now, let's substitute the given values:
Q = (40.0 N/C) * (0.250 m)^2 / (8.99 x 10^9 N m^2/C^2)
Calculating this:
Q = (40.0 * 0.0625) / (8.99 x 10^9)
Q = 2.5 x 10^-9 C
Therefore, the charge of object A is 2.5 nanocoulombs (nC).