Charges of 4 µC are located at x = 0, y = 2.0 m and at x = 0, y = -2.0 m. Charges Q are located at x = 4.0 m, y = 2.0 m and at x = 4.0 m, y = -2.0 m. The electric field at x = 0, y = 0 is (4 ✕ 103 N/C) i. Determine Q.
at (0,0) the fields from the first two charges cancel each other.
to get a field at (0,0) in the y direction the x components of charges Q must cancel.
Therefore one must be +Q and the other -Q
If the field is up (+ i direction)
the one below the axis is +Q and the one above the axis is -Q
each contributed half of the E field
distance = d = sqrt (20) = 2 sqrt 5
so d^2 = 20
E = 2 *10^3 = k Q/20
so Q = (2*10^3*20)/(8.99*10^9) ?
i calculated that and it doesnt give me the correct answer
To determine the charge Q, we need to use the concept of electric field and Coulomb's law.
The electric field is a property of charged objects that exerts a force on other charged objects. The electric field at a point is defined as the force experienced by a unit positive charge placed at that point.
In this problem, we are given the value of the electric field at point (0, 0) as (4 ✕ 10^3 N/C)i. This means that the electric field at that point is directed along the x-axis (i.e., horizontal) with a magnitude of 4 ✕ 10^3 N/C.
To find the charge Q, we can use Coulomb's law, which relates the electric force between two charges to their magnitudes and the distance between them.
Coulomb's law states that the electric field E between two charges q1 and q2 at a distance r apart is given by the equation:
E = (k |q1| |q2|) / r^2
Where:
E is the electric field
k is the electrostatic constant (k ≈ 9 ✕ 10^9 Nm^2/C^2)
|q1| and |q2| are the magnitudes of the charges
r is the distance between the charges
In our case, we have two charges of 4 µC each (|q1| = |q2| = 4 ✕ 10^-6 C) at positions (0, 2) and (4, 2) meters respectively. The distance between them is 4 meters (r = 4 m).
Using Coulomb's law, we can express the electric field at point (0, 0) as the sum of the electric field contributions from both charges:
E = E1 + E2
Since the electric field is a vector quantity, the x-components (in this case, horizontal components) of the electric fields add up, while the y-components cancel out.
Therefore, we have:
4 ✕ 10^3 N/C = (k |q1| / r^2) + (k |q2| / r^2)
Plugging in the values of k, |q1|, |q2|, and r:
4 ✕ 10^3 = (9 ✕ 10^9 Nm^2/C^2) (4 ✕ 10^-6 C) / (4 m)^2
4 ✕ 10^3 = (9 ✕ 10^3 Nm^2/C^2) (4) / (4 m)^2
4 ✕ 10^3 = (9 ✕ 10^3 Nm^2/C^2) / (4 m^2)
Now we can solve for Q:
Q = (4 ✕ 10^3) (4 m^2) / (9 ✕ 10^3 Nm^2/C^2)
Q = (16 ✕ 10^3 m^2/N) / (9 ✕ 10^3 m^2/C)
Q = 16/9 C
So, the charge Q is approximately 1.78 C.