Solving for Electric Force: A 15cm Ring, 20uC Charge

In summary, the task was to calculate the force on a -3 micro coulombs charge located on the x-axis at x=5cm, in the presence of a ring with a 15cm radius and a uniform charge of 20 micro coulombs in the yz-plane. This was done using the equations E=kQx/(R^2 + x^2)^(3/2) and F=QE, and the final answer was determined to be -6.83 N.
  • #1
huntingrdr
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0

Homework Statement


A ring with a 15cm radius and with a uniform charge of 20 micro coulombs is in the yz-plane with the origin at its center. What is the force on a -3 micro coulombs charge on the x-axis at x=5cm?


Homework Equations



E=kQx/(R^2 + x^2)^(3/2)

F=QE

The Attempt at a Solution



I tried to plug in the numbers for the above equations and can't seem to get the right answer. I first solved for E and then multiplied it to Q which is -3*10^(-6) C...correct?
The answer should be -6.83 N, but I can't seem to figure it out.
 
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  • #2
I got -6.8 N when I used your numbers. Check your math, check your unit conversions for the distances and charges, and make sure you're using the correct Q for the electric field.
 
  • #3
Ahh...I made a mistake on my calculator. Thanks for the help.
 

FAQ: Solving for Electric Force: A 15cm Ring, 20uC Charge

How can I calculate the electric force on a 15cm ring with a 20uC charge?

To calculate the electric force, you will need to use the equation F = k * (q1 * q2) / r^2, where F is the electric force, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between the charges. In this case, q1 is 20uC and r is 15cm (or 0.15m).

What is the direction of the electric force on the 15cm ring?

The direction of the electric force depends on the sign of the charges. If both charges are positive or negative, the force will be repulsive and the direction will be away from the other charge. If one charge is positive and the other is negative, the force will be attractive and the direction will be towards the other charge.

Can I use this equation to calculate the electric force on any charged object?

Yes, this equation can be used to calculate the electric force on any charged object as long as the distance between the charges is known. However, if the object is not a point charge (meaning it has a non-zero size), the equation may need to be modified accordingly.

How does the electric force change if the distance between the charges is doubled?

According to the equation F = k * (q1 * q2) / r^2, the electric force is inversely proportional to the square of the distance between the charges. This means that if the distance is doubled, the force will decrease by a factor of 4. Similarly, if the distance is halved, the force will increase by a factor of 4.

Are there any other factors that can affect the electric force on a charged object?

Other factors that can affect the electric force include the magnitude of the charges, the medium in which the charges are located, and the presence of other charged objects. These factors can change the value and direction of the electric force.

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