# Physic 2 Need Help Please?

Two positive charges q1= q2 = 2.0 uC are located at x = 0, y = 0.30 m and x = 0, y= -0.30 m, respectively. Third point charge Q = 4.0 uC is located at x = 0.40 m , y = 0.What is the net force ((a)magnitude and (b)direction) on charge q1 exerted by the other two charges?

A) Find the Magnitude (| F–>|) N. Express answer using two significant numbers.

B) Find the Direction (θ) degree counterclockwise from the x direction.

• electron1
9 days ago

Two positive charges q1= q2 = 2.0 uC are located at x = 0, y = 0.30 m and x = 0, y= -0.30 m, respectively. Third point charge Q = 4.0 uC is located at x = 0.40 m, y = 0.What is the net force ((a)magnitude and (b)direction) on charge q1 exerted by the other two charges?

F = (k * q1 * q2) ÷ d^2

k = 9 * 10^9

q1 = q2 = 2 * 10^-6 C , q3 = 4 * 10^-6 C

Since all the charges are positive, the force, between any two charges, will cause the objects to repel each other.

To determine the force, you need to know the distance from q1 to each of the other charges. If you sketch a graph of the 3 points, you can see where they are with respect to each other. q1 is 0.3 m above the origin, and q2 is 0.3 m below the origin. q3 is 0.4 m to the right of the origin. If you draw a line from q1 to q3, you will see a right triangle. The base of the right triangle is 0.4 m and the height is 0.3 m.

Position of q1 = (0, +0.3)

Position of q2 = (0, -0.3)

Position of q3 = (0.4. 0)

Go to the website below to see the right triangle.

http://emweb.unl.edu/Math/mathweb/trigonom/Image28…

In this right triangle, the 90˚ angle is at the origin. q2 is 0.3 m above the origin, and q3 is 0.4 m to the left of origin.

In your problem, 1 is 0.3 m above the origin, q2 is 0.3 m below the origin, and q3 is 0.4 m to the right of origin.

The force on q1 due to q2, will cause q1 to accelerate in the +y direction. The force on q1 due to q3, will cause q1 to accelerate in the +y direction and the negative x direction.

Use the equation below to determine the distance between two points.

d = [x1 – x2)^2 + (y1 – y2)^2]^0.5

For q2, d = [(0 – 0 )^2 + (3 – -3)^2]^0.5 = 0.6 m

For q3, d = [(0 – 0.4)^2 + (0.3 – 0)^2]^0.5 = 0.5 m

Force on q1 due to q2 = (9 * 10^9 * 2 * 10^-6 * 2 * 10^-6) ÷ 0.6^2 = 0.1 N

Since q1 is above q2, this force will cause q1 to accelerate in the positive y direction.

Force on q1 due to q3 = (9 * 10^9 * 2 * 10^-6 * 4 * 10^-6) ÷ 0.5^2 = 0.288 N

Since q3 is directly below q2, the force on q1 due to q2 is in the +y direction. The force on q1 due to q3 has a positive y component and a negative x component. The magnitude of each component is dependent on the angle between the x axis and the line from q1 to q3; which is the hypotenuse of the right triangle. Since q1 is to the left of q3, the force will cause q1 to accelerate to the left, in the negative x direction. Since q1 is above q3, the force will cause q1 to accelerate in the positive y direction.

x component = -F * cos θ, y component = F * sin θ

Now you need to determine the sine and cosine of the angle between the hypotenuse of the right triangle and the x axis.

you will see the right triangle. The height is 0.3 m, and the base is 0.4 m. This is a 3 – 4 – 5 right triangle. So the length of the hypotenuse is 0.5 m.

Cos θ = base ÷ hypotenuse = 0.4 ÷ 0.5 = 0.8

x component = F * 0.8

x component = -0.288 * 0.8 = -0.2304 N

Sin θ = height ÷ hypotenuse = 0.3 ÷ 0.5 = 0.6

y component = F * 0.6

y component = 0.288 * 0.6 = +0.1728 N

To determine the net force on q1, we need to determine the sum of the y components.

Sum of y components = 0.1 + 0.1728 = 0.2728 N

Net force = (-0.2304, 0.2728)

On graph paper, this point is in the 3rd quadrant. The net force vector is above the negative x axis. The tangent of the angle above the negative x axis = 0.2728 ÷ 0.2304. The angle is approximately 49.8˚ above the negative x axis. The problem asks for the degree counterclockwise from the +x direction.

The negative x direction is 180˚ counterclockwise from the +x direction. 180 – 49.8 = 130.2

The angle from the +x direction = 130.2˚

Net force on q1 = [(x component)^2 + (sum of y components)^2]^0.5

Net force = [(-0.2304)^2 + (0.2728)^2]^0.5 = 0.357 N