a charge q1=5.00nc is placed at the origin of an xy-coordinate system ,and a charge q2=-2.00nc is placed on the positive x-axis at x=4.00cm (a.) if a third charge q3=6.00nc is now placed at the point x=4cm, y=3.00cm. find the x and y components of the total force exerted on this charge by the other two. (b.) find the magnitude and direction of this force.
To find the components of the total force exerted on charge q3 by the other two charges:
(a.) First, let's find the x-component of the force.
The force exerted by charge q1 on charge q3 can be calculated using Coulomb's Law:
F1 = k * q1 * q3 / r1^2
where k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), q1 is the charge of object 1 (5.00 nC), q3 is the charge of object 3 (6.00 nC), and r1 is the distance between object 1 and object 3.
Since q1 is at the origin (0, 0), and q3 is located at (x = 4 cm, y = 3 cm), the distance r1 can be calculated using the Pythagorean theorem:
r1 = sqrt(4^2 + 3^2)
Substituting the values, we have:
r1 = sqrt(16 + 9) = sqrt(25) = 5 cm
Now we can calculate the x-component of the force exerted by q1 on q3:
F1x = F1 * cos(theta1)
where theta1 is the angle between the x-axis and the line connecting q1 and q3. Since q1 is at the origin, the angle theta1 is the angle between the line connecting q1 to q3 and the positive x-axis.
cos(theta1) = adjacent / hypotenuse
cos(theta1) = 4 cm / 5 cm
Substituting the values, we have:
F1x = F1 * (4 / 5)
Similarly, the x-component of the force exerted by q2 on q3 can be calculated in a similar way. The only difference is that the distance r2 is the distance between q2 (at x = 4 cm, y = 0) and q3.
r2 = sqrt((4 - 4)^2 + (3 - 0)^2) = sqrt(9) = 3 cm
The angle theta2 is the angle between the line connecting q2 to q3 and the positive x-axis. Since q2 is positioned at x = 4 cm, the angle theta2 is 0 degrees.
Therefore, F2x = F2 * cos(theta2) = F2 * cos(0) = F2, as cos(0) equals 1.
(b.) To find the magnitude and direction of the total force exerted on q3, we need to sum up the x-components and y-components of the forces:
F_total_x = F1x + F2x
F_total_y = F1y + F2y
The magnitude of the total force is given by:
F_total = sqrt(F_total_x^2 + F_total_y^2)
The direction of the force can be found using the atan2 function:
angle = atan2(F_total_y, F_total_x)
Now let's calculate the values:
1. Calculate F1x:
F1x = (8.99 x 10^9 N m^2/C^2) * (5.00 x 10^-9 C) * (6.00 x 10^-9 C) / (5 cm)^2
2. Calculate F2x:
F2x = (8.99 x 10^9 N m^2/C^2) * (-2.00 x 10^-9 C) * (6.00 x 10^-9 C) / (3 cm)^2
3. Calculate F_total_x:
F_total_x = F1x + F2x
4. Find F1y and F2y:
Both F1y and F2y will be 0 since the charges q1 and q2 are positioned on the x-axis. So F1y = 0 and F2y = 0.
5. Calculate F_total_y:
F_total_y = F1y + F2y = 0
6. Calculate the magnitude of the total force:
F_total = sqrt(F_total_x^2 + F_total_y^2)
7. Calculate the direction of the total force:
angle = atan2(F_total_y, F_total_x)
By following these steps and performing the calculations, you will find the x and y components of the total force, as well as the magnitude and direction of the force exerted on charge q3.
To find the x and y components of the total force exerted on charge q3 by charges q1 and q2, we can use Coulomb's Law, which states that the force between two charges is given by:
F = (k * |q1 * q2|) / r^2
where F is the force, k is the electrostatic constant (k = 8.99 * 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges. However, we also need to consider the direction of the force.
(a) Calculating the x and y components of the force:
1. First, let's calculate the force exerted by q1 on q3:
Given:
- q1 = 5.00 nC
- q3 = 6.00 nC
The distance r between the charges is given by the Pythagorean theorem:
r = √[(x2 - x1)^2 + (y2 - y1)^2]
Substituting the values:
r = √[(4 cm - 0 cm)^2 + (3 cm - 0 cm)^2]
r = √[16 cm^2 + 9 cm^2]
r = √[25 cm^2]
r = 5 cm
Now we can calculate the x and y components of the force exerted by q1 on q3 using Coulomb's Law:
Fx1 = (k * |q1 * q3|) / r^2
Fx1 = (8.99 * 10^9 N m^2/C^2 * |5.00 * 6.00 * 10^-9 C|) / (5 * 10^-2 m)^2
Simplifying the equation:
Fx1 = (8.99 * 5.00 * 6.00) / (25 * 10^2)
Fx1 = 1.798 N
Since the positive x-axis is directed to the right, the x-component of the force exerted by q1 on q3 is 1.798 N to the right.
Fy1 = 0 N (There is no y-components of force since q1 is placed on the origin, so there is no vertical component.)
2. Second, let's calculate the force exerted by q2 on q3:
Given:
- q2 = -2.00 nC
- q3 = 6.00 nC
Using Coulomb's Law again, we get:
Fx2 = (k * |q2 * q3|) / r^2
Fx2 = (8.99 * 10^9 N m^2/C^2 * |-2.00 * 6.00 * 10^-9 C|) / (5 * 10^-2 m)^2
Simplifying the equation:
Fx2 = -1.799 N
The negative sign indicates that the force exerted by q2 on q3 is in the opposite direction of the positive x-axis.
Fy2 = 0 N (There is no y-component of force since q2 lies on the positive x-axis, so there is no vertical component.)
Therefore, the x and y components of the total force exerted on q3 by q1 and q2 are:
Fx_total = Fx1 + Fx2 = 1.798 N + (-1.799 N) = -0.001 N
Fy_total = Fy1 + Fy2 = 0 N + 0 N = 0 N
(b) Finding the magnitude and direction of the force:
The magnitude of the force is given by the magnitude of the vector with x and y components:
|F_total| = √(Fx_total^2 + Fy_total^2)
|F_total| = √((-0.001 N)^2 + (0 N)^2)
|F_total| = √(0.000001 N^2)
|F_total| = 0.001 N
The direction of the force can be determined using the tangent of the angle:
θ = atan(Fy_total / Fx_total)
θ = atan(0 N / -0.001 N)
θ = atan(0)
θ = 0 radians (or 0 degrees since it's a right angle)
Therefore, the magnitude of the force is 0.001 N and the direction of the force is at an angle of 0 degrees with the negative x-axis.
1. First draw a picture of the situation, so that you can see where these charges are located on the Cartesian Plane, and the angles involved.
2. Use Coulomb's Law:
Force of q1 on q3 (it pushes q3 away)
F1= (kq1q3)/(r)^2
Force of q2 on q3 (it pulls on q3)
F2 = (kq2q3)/(r)^2
3. Draw a FBD of these two forces acting on q3. Sum forces in the x direction and then in the y direction to get a single force for each. Find the resultant magnitude and direction of the x,y-direction forces using trig, as usual.