three charges are fixed to an x, y coordinate system. a charge of +18 µC is on the y axis at y= +3.0 degree m. a charge of -12 µC is at the origin. last, a charge of +45 µC is on the x axis at x= +3.0 m. determine the magnitude and direction of the net electrostatic force on the charge at x= +3.0 m. specify the direction relative to the -x axis.
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To determine the net electrostatic force on the charge at x= +3.0 m, we need to calculate the individual forces between the charges and then add them up vectorially.
First, let's determine the force between the charge at x= +3.0 m and the charge at the origin.
The formula for the electrostatic force is given by Coulomb's Law:
F = k * (|q1 * q2| / r^2)
Where:
- F is the force between the charges
- k is the electrostatic constant and its value is 9 x 10^9 N.m^2/C^2
- q1 and q2 are the charges
- r is the distance between the charges
For the force between the charge at x= +3.0 m and the charge at the origin, q1 = +45 µC and q2 = -12 µC. The distance between them is the x-coordinate of the charge at x= +3.0 m, which is 3.0 m.
Using Coulomb's Law, the force between these charges is:
F1 = k * (|q1 * q2| / r^2)
= 9 x 10^9 N.m^2/C^2 * [(45 x 10^-6 C) * (-12 x 10^-6 C) / (3.0 m)^2]
Calculating this, we get:
F1 = -0.54 N
The negative sign indicates that the force is attractive.
Now, let's determine the force between the charge at x= +3.0 m and the charge on the y-axis at y= +3.0 m.
For the force between the charge at x= +3.0 m and the charge on the y-axis, q1 = +45 µC and q2 = +18 µC. The distance between them is the y-coordinate of the charge on the y-axis, which is 3.0 m.
Using Coulomb's Law, the force between these charges is:
F2 = k * (|q1 * q2| / r^2)
= 9 x 10^9 N.m^2/C^2 * [(45 x 10^-6 C) * (18 x 10^-6 C) / (3.0 m)^2]
Calculating this, we get:
F2 = 0.18 N
The positive sign indicates that the force is repulsive.
Now, let's find the net force by adding the individual forces vectorially. Since the forces have different directions, we need to add them as vectors.
Net Force = F1 + F2
In terms of magnitude and direction, the net force is given by:
Magnitude = sqrt(F1^2 + F2^2)
Direction = Arctan(F2 / F1)
Calculating this, we get:
Magnitude = sqrt((-0.54 N)^2 + (0.18 N)^2) = 0.57 N (rounded to two decimal places)
Direction = Arctan(0.18 N / -0.54 N) = -18.44°
Therefore, the magnitude of the net electrostatic force on the charge at x= +3.0 m is 0.57 N, and the direction relative to the -x axis is approximately 18.44° to the right of the -x axis.
To find the magnitude and direction of the net electrostatic force on the charge at x = +3.0 m, we need to calculate the individual forces between the charge at x = +3.0 m and the other charges, and then combine them vectorially.
Let's first calculate the magnitude of the force between the charge at x = +3.0 m and the charge at the origin.
The formula to calculate the magnitude of the electrostatic force between two charges is given by Coulomb's Law:
F = (k * |q1 * q2|) / r^2
Where:
F is the magnitude of the force
k is the electrostatic constant, approximately 9.0 x 10^9 Nm^2/C^2
q1 and q2 are the charges
r is the distance between the charges
In this case, q1 is +45 µC, q2 is -12 µC, and the distance r is 3.0 m (the same as the distance between the charges).
Plugging in these values into the formula, we get:
F1 = (9.0 x 10^9 Nm^2/C^2 * (45 x 10^-6 C) * (12 x 10^-6 C)) / (3.0 m)^2
Calculating this expression gives us:
F1 ≈ 9.6 x 10^4 N
The direction of this force is towards the charge at the origin, which is in the negative x-direction.
Next, let's calculate the magnitude of the force between the charge at x = +3.0 m and the charge on the y-axis at y = +3.0 m.
Since both charges are positioned symmetrically with respect to the x-axis, the force between them will be equal in magnitude but opposite in direction to the previous force. So, the magnitude of this force is also 9.6 x 10^4 N, but the direction is now towards the positive x-direction.
Now, we can combine these two forces vectorially to find the net force.
Since the forces are equal in magnitude and opposite in direction, their net force will be zero. Therefore, the magnitude of the net force is 0 N.
As for the direction, since the net force is zero, we can say that the direction is ambiguous or indeterminate in this case.
To summarize:
Magnitude of the net force on the charge at x = +3.0 m: 0 N
Direction of the net force: Ambiguous, as the net force is zero.