Coulomb's Law: How to Calculate Electric Force in a Right Triangle

• kavamo
In summary: I am unsure of what to do next (how to proceed). Please give example. Thanks.In summary, the conversation discusses finding the electric force on a point charge due to two other fixed charges in a right triangle. The formula F=kq1q2/r^2 is used, with k=8.99x10^9. After some calculations, the numbers seem large and it is suggested to check for proper unit conversion. Once the units are converted correctly, the next step is to use trigonometry to find the magnitude and direction of the electric force by representing it as a single vector.
kavamo
Coulombs Law--what now?

Homework Statement

Three point charges are fixed in place in a right triangle. What is the electric force on the q = -0.64 µC charge due to the other two charges? (Let Q1 = +0.71 µC and Q2 = +1.1 µC.)

Find:

magnitude N

and

direction ° above the positive x-axis

F=kq1q2 / r^2

K=8.99 x 10^9

The Attempt at a Solution

The diagram from the book shows a right triangle where q is the 90 degrees vertex along X-axis. the hypotenuse = 10 cm; r (y)=8.0 cm; and between q & Q2 =6 cm.

I have made free body diagrams and know: q = -0.64 µC; on the y-axis Q1= +0.71 µC and Q2 = +1.1 µC on the X-axis

plugging the info into the formulas I get:

F (y)= (8.99x10^9)(0.71)(0.64) / 8^2 = 6382900

F (x)= (8.99x10^9)(1.1)(0.64) / 6^2 = 175804444.4

Don't these numbers seem rather big? Did I miss a conversion somewhere?

I am also unsure of what to do next.

Those numbers are definitely plausible. You have to remember that k = 9e9. So if those numbers are correct, what you want to do is draw the vectors associated with them and use cosine law most likely to find the summation of those vectors.

kavamo said:

Homework Statement

Three point charges are fixed in place in a right triangle. What is the electric force on the q = -0.64 µC charge due to the other two charges? (Let Q1 = +0.71 µC and Q2 = +1.1 µC.)

Find:

magnitude N

and

direction ° above the positive x-axis

F=kq1q2 / r^2

K=8.99 x 10^9

The Attempt at a Solution

The diagram from the book shows a right triangle where q is the 90 degrees vertex along X-axis. the hypotenuse = 10 cm; r (y)=8.0 cm; and between q & Q2 =6 cm.

I have made free body diagrams and know: q = -0.64 µC; on the y-axis Q1= +0.71 µC and Q2 = +1.1 µC on the X-axis

plugging the info into the formulas I get:

F (y)= (8.99x10^9)(0.71)(0.64) / 8^2 = 6382900

F (x)= (8.99x10^9)(1.1)(0.64) / 6^2 = 175804444.4

Don't these numbers seem rather big? Did I miss a conversion somewhere?

I am also unsure of what to do next.

These numbers seem big, because you're not converting any of the units to the standard SI! Convert the microCoulombs to Coulombs and the centimeters to meters. Once you correct your order of magnitude by proper unit conversion, your next step should be trigonometric. Think of the x and y components as sides of a triangle with the hypotenuse equal to the magnitude.

Last edited:

thanks I'll try it again.

xcvxcvvc said:
These numbers seem big, because you're not converting any of the units to the standard SI! Convert the microCoulombs to Coulombs and the centimeters to meters. Once you correct your order of magnitude by proper unit conversion, your next step should be trigonometric. Think of the x and y components as sides of a triangle with the hypotenuse equal to the magnitude.

Hi. After converting as per your advice, I have the following numbers:

F (Q1q) = (8.99 x 10^9)(7.1 x 10^-5)(6.4 x 10^-5) / .08^2 = 6382.9

F (Q2q) = (8.99 x 10^9)(1.1 x 10^-4)(6.4 x 10^-5) / .06^2 = 17580.44444

Please check this for accuracy (I'm new to the calculator apps--it's likely I made errors).

If conversions are correct--what is my next step? Please give example. Thanks.

kavamo said:
Hi. After converting as per your advice, I have the following numbers:

F (Q1q) = (8.99 x 10^9)(7.1 x 10^-5)(6.4 x 10^-5) / .08^2 = 6382.9

F (Q2q) = (8.99 x 10^9)(1.1 x 10^-4)(6.4 x 10^-5) / .06^2 = 17580.44444

Please check this for accuracy (I'm new to the calculator apps--it's likely I made errors).

If conversions are correct--what is my next step? Please give example. Thanks.

Your units are still converted incorrectly.
$$\frac{.71\mu C}{1}*\frac{1C}{10^6\mu C}=710x10^{-9}\neq 7.1x10^{-5} C$$

I believe your other charge conversions are faulty too.

http://img682.imageshack.us/img682/6978/70405368.jpg
Assuming you did your math right and both your forces are positive, this is a picture of what you're trying to find in your next step. Instead of showing the answer as two components - one vector at 0 degrees and the other at 90 degrees - you do vector addition to represent your answer as one vector. This single vector means you need a magnitude and a direction.

To find the magnitude, use 90 degree trigonometry. To find the angle, do the same. EDIT: I also just realized I typed the two components in reverse. The x component is supposed to be on the x-axis, and the y-component is supposed to be on the y-axis.

Last edited by a moderator:

xcvxcvvc said:
Your units are still converted incorrectly.
$$\frac{.71\mu C}{1}*\frac{1C}{10^6\mu C}=710x10^{-9}\neq 7.1x10^{-5} C$$

I believe your other charge conversions are faulty too.]

Thank you.

New conversions:

F(x)= (8.99x10^9) x [(1.1x10^-6)(6.4x10^-7) / (.06^2)] = 1.758044444

F(y)= (8.99x10^9) x [(7.1x10^-9)(6.4x10^-7) / (.08^2)] = .0063829

Are these calculations correct?

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law of electrostatics that describes the force between two charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. How is Coulomb's Law calculated?

Coulomb's Law is calculated using the equation F = k(q1q2)/r2, where F is the force between the two particles, k is a constant, q1 and q2 are the charges of the two particles, and r is the distance between them.

3. What is the unit of measurement for Coulomb's Law?

The unit of measurement for charge in Coulomb's Law is the Coulomb (C), and the unit of measurement for distance is meters (m). Therefore, the unit for the force in Coulomb's Law is Newtons (N).

4. How does Coulomb's Law relate to electric fields?

Coulomb's Law is a mathematical representation of the electric field, which is the force per unit charge experienced by a charged particle. The electric field is equal to the force between two charged particles divided by the charge of one of the particles.

5. What is the importance of Coulomb's Law in science and technology?

Coulomb's Law is a fundamental law of physics that has many applications in science and technology. It is used to understand the behavior of charged particles in electric fields, which is essential in fields such as electronics, telecommunications, and electromagnetism. It also helps in understanding the properties of materials and the behavior of atoms and molecules.

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