Three point charges are fixed in place in a right triangle. What is the electric force on the q = -0.63 µC charge due to the other two charges? (Let Q1 = +0.83 µC and Q2 = +0.9 µC.)
what is the magnitude N, and direction in degrees ° above the positive x-axis?
Let's start by drawing a diagram of the overall setup. Place Q1 at the origin (0,0), Q2 right from Q1 along the x-axis, and q on top of the triangle above the origin. Let the distance between Q1 and q be a and between Q1 and Q2 be b.
We can use Coulomb's Law to find the magnitudes of the forces between the charges. The electric force between two charges is given by:
F = k * |q1 * q2| / r^2
where k is the electrostatic constant (approximately 8.99 * 10^9 N(m/C)^2), q1 and q2 are the magnitudes of the point charges, and r is the distance between them.
First, we'll find the magnitude of the force between Q1 and q:
F_Q1q = k * |Q1 * q| / a^2
Then, we'll find the magnitude of the force between Q2 and q:
F_Q2q = k * |Q2 * q| / b^2
Next, we'll find the horizontal and vertical components of these forces.
The force between Q1 and q has a horizontal component F_Q1qx and a vertical component F_Q1qy. The horizontal component is perpendicular to the right triangle and is given by:
F_Q1qx = F_Q1q
The vertical component points straight up and is given by:
F_Q1qy = 0
The force between Q2 and q has a horizontal component F_Q2qx and a vertical component F_Q2qy. The horizontal component is perpendicular to the right triangle and is given by:
F_Q2qx = F_Q2q * (b / sqrt(a^2 + b^2))
The vertical component points straight up and is given by:
F_Q2qy = F_Q2q * (a / sqrt(a^2 + b^2))
Now we can add the horizontal and vertical components to find the total force.
F_x = F_Q1qx - F_Q2qx = F_Q1q - F_Q2q * (b / sqrt(a^2 + b^2))
F_y = F_Q1qy + F_Q2qy = F_Q2q * (a / sqrt(a^2 + b^2))
Next, we'll find the magnitude and direction of the total force.
magnitude = sqrt(F_x^2 + F_y^2)
direction = arctan(F_y / F_x)
Plugging in the values for a, b, Q1, Q2, and q from the problem, we get:
F_Q1q = 8.99 * 10^9 * (0.83 * 10^(-6) * -0.63 * 10^(-6)) / a^2
F_Q2q = 8.99 * 10^9 * (0.9 * 10^(-6) * -0.63 * 10^(-6)) / b^2
magnitude = sqrt((F_Q1q - F_Q2q * (b / sqrt(a^2 + b^2)))^2 + (F_Q2q * (a / sqrt(a^2 + b^2)))^2)
direction = arctan(F_Q2q * (a / sqrt(a^2 + b^2)) / (F_Q1q - F_Q2q * (b / sqrt(a^2 + b^2))))
Note that the values for a and b are not given, so we cannot find the magnitude and direction of the total force numerically. However, you can plug in the known values for Q1, Q2, and q into these expressions to find the formulas for the magnitude and direction of the total force on q due to the other two charges.
To find the electric force on the q = -0.63 µC charge due to the other two charges Q1 and Q2, we can use the equation for the electric force between two charges:
F = k * |q1| * |q2| / r^2
Where:
- F is the magnitude of the electric force
- k is the electrostatic constant (k = 9 × 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 Q1 = +0.83 µC, Q2 = +0.9 µC, and q = -0.63 µC.
First, let's find the distances between the charges. Since the charges are fixed in a right triangle, we can use the Pythagorean theorem to find the distance between Q1 and the q charge:
Distance between Q1 and q = sqrt((1 unit)^2 + (2 units)^2) = sqrt(5) units
Similarly, the distance between Q2 and q is the same: sqrt(5) units.
Now, we can calculate the electric force exerted by Q1 on q:
F1 = k * |q1| * |q| / r^2
= (9 × 10^9 N m^2/C^2) * (0.83 µC) * (0.63 µC) / (5 units)^2
= (9 × 10^9) * (0.83) * (0.63) / 5^2
= (9 × 0.83 × 0.63 × 10^9) / 5^2
= 3.447 N
The direction of this force can be found using trigonometry. The angle above the positive x-axis can be calculated as:
angle1 = atan(2/1)
= atan(2)
≈ 63.43°
Similarly, the electric force exerted by Q2 on q is:
F2 = k * |q2| * |q| / r^2
= (9 × 10^9 N m^2/C^2) * (0.9 µC) * (0.63 µC) / (5 units)^2
= (9 × 0.9 × 0.63 × 10^9) / 5^2
= 3.671 N
The angle above the positive x-axis for this force is:
angle2 = atan(1/2)
= atan(0.5)
≈ 26.57°
To find the total magnitude of the electric force on q, we can use the principle of superposition:
F_total = sqrt(F1^2 + F2^2)
= sqrt((3.447 N)^2 + (3.671 N)^2)
≈ 4.875 N
The direction of the total force can be found using the angle between the force components:
angle_total = atan(F2 / F1)
= atan(3.671 N / 3.447 N)
≈ 44.62°
Therefore, the magnitude of the electric force on the q = -0.63 µC charge is approximately 4.875 N, and it is inclined at an angle of about 44.62° above the positive x-axis.
To find the electric force on the q = -0.63 µC charge due to the other two charges, we can use Coulomb's Law. Coulomb's Law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The equation for Coulomb's Law is:
F = k * (|q1 * q2|) / r^2
Where:
F is the electric force between the charges
k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2)
q1 and q2 are the charges
r is the distance between the charges
In this case, we have three charges arranged in a right triangle. Let's label the charges as follows:
Q1 = +0.83 µC
Q2 = +0.9 µC
q = -0.63 µC
Since the charges Q1 and Q2 are fixed in place, we only need to consider the force between Q1 and q, and the force between Q2 and q.
To find the magnitude of the force, we can plug the values into the formula:
F1 = k * (|Q1 * q|) / r1^2
F2 = k * (|Q2 * q|) / r2^2
We need to find r1 and r2, which are the distances between the charges. In a right triangle, we can use the Pythagorean theorem to find the distances:
r1 = sqrt(d^2 + h^2)
r2 = sqrt(b^2 + h^2)
Where:
d is the horizontal distance between Q1 and q
b is the horizontal distance between Q2 and q
h is the vertical distance between the charges
To find the direction of the force in degrees above the positive x-axis, we can use the inverse tangent function:
θ1 = arctan(h / d)
θ2 = arctan(h / b)
Finally, we can calculate the net force by adding the vector components of F1 and F2, and find the magnitude and direction of that net force using trigonometry.
By following these steps, you can determine the magnitude and direction of the electric force on the q = -0.63 µC charge due to the other two charges.