HELP! NEED ASAP :( Two equal positive charges q1 = q2 = 2.0 mC are located at x = 0, y = 0.30 m and...

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HELP! NEED ASAP 🙁
Two equal positive charges q1 = q2 = 2.0 mC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. What are the magnitude and direction of the total electric force that q1 and q2 exert on a third charge Q = 4.0 mC at x = 0.40 m, y = 0?

Four charges are arranged in a square with one side of the square measuring 0.2m. From upper left going clockwise, q1 = 2µC; q2 = -4µC; q3 = 6µC; and q4 = -8µC. Solve for the amount of electric force experienced by q4 due to the other charges.

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Two equal positive charges q1 = q2 = 2.0 mC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. What are the magnitude and direction of the total electric force that q1 and q2 exert on a third charge Q = 4.0 mC at x = 0.40 m, y = 0?

Four charges are arranged in a square with one side of the square measuring 0.2m. From upper left going clockwise, q1 = 2µC; q2 = -4µC; q3 = 6µC; and q4 = -8µC. Solve for the amount of electric force experienced by q4 due to the other charges.

Answer:

The magnitude of the total electric field is 1.152 x [tex]10^8[/tex] N/C and its direction is to the right toward the the third charge.

Explanation:

in our case, we are to solve for the total electric field and direction two positive, identical charges towards another charge.

Here, we are to solve first the total force of the charges with the third charge. to solve this, we will apply the concept of Coulomb’s law.

1. For the force between charges, we will use the formula:

[tex]F=9×10^9frac{N.m^2}{C^2} frac{qq’}{r^2}[/tex]

2. And for the electric field, we will use the formula:

E = F/q’

For the given information

q1 = q2 = 2.0 mC = [tex]2×10^{-3}C[/tex]

q3 = 4.0 mC = [tex]4×10^{-3}C[/tex]

Solving the problem

1. Let us solve first the total force of the charges q1 and q2 on the third charge, q3.

It is easy to analyze the problem by using a vector diagram of forces. I have provided the diagram below.

Let us solve first the distance between q1 and q3 which is also the same distance of q2 and q3 using Pythagorean theorem.

r = [tex]sqrt{(0.3m)^2+(0.4m)^2}[/tex]

r = 0.5m

Now, let us use the formula for force and then substitute the given information, we have:

[tex]F=9×10^9frac{N.m^2}{C^2} frac{qq’}{r^2}[/tex]

[tex]F=9×10^9frac{N.m^2}{C^2} frac{(2×10^{-3}C)(4×10^{-3}C)}{(0.5m)^2}[/tex]

F = 288,000 N

Now, there are x and y components of q1 and q2 towards q3. Let us solve for them, we have:

Fx = F cos θ                                Fy = F sin θ                              

Fx = 288,000 [tex](frac{0.4}{0.5} )[/tex]                     Fy = 288,000  [tex](frac{0.3}{0.5} )[/tex]  

Fx = 230,400 N                         Fy = 172,800 N

For the total force, let us solve for the sum for the x and y components

x-component:

∑Fx = Fx1 + Fx1

where Fx1  is the x component of q1 and Fx2 is the x component of q2. Since they are identical and both of them goes to the right (positive), we add them

∑Fx = 230,400 N + 230,400 N

∑Fx = 460,800 N

y-component:

∑Fy = – Fy1 + Fy2

where   Fy1 is the y-component of q1 and Fy2 is the y-component of q2. Referring to the diagram below, Fy1 is negative because its direction is downward while Fy2 is positive because its direction is upward.

∑Fy = – Fy1 + Fy2 = 0

Since the y-component force is identical and opposite in direction, therefore the summation of force is zero.

Therefore, solving for the total force:

Ft = [tex]sqrt{(Sigma Fx)^2+(Sigma Fy)^2 }[/tex]

Ft = 460,800 N

2. Now, we are ready to solve for the total electric field on q3. Let us use the formula for electric field and substitute the given information, we have:

E = F/q’   or

Et = Ft / q3

Et = 460,800 N / 4 x [tex]10^{-3}[/tex] C

Et = 1.152 x [tex]10^{8}[/tex] N/C

Therefore, the magnitude of the total electric field is 1.152 x [tex]10^{8}[/tex] N/C and its direction is the right.

Remember:

Take note that the when a charge is positive, the direction of the force is the same direction of the electric field and when the charge is negative, the direction of the force goes on the opposite direction.

The direction between charges is from positive towards the negative charge or the direction always start from the positive charge going to another charge.

Now, you can practice answering on the second problem.

To learn more about this topic, just click on the following links:

• Definition of electric field

       https://brainly.ph/question/106082

• Additional example

       https://brainly.ph/question/2468653

       https://brainly.ph/question/2484031

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