Calculating Electric Field Strength and Acceleration in a Coulombic System

In summary, a 40 Micro Coluomb test charge is placed in an electric field produced by a 5 Coloumb charge, experiencing a 6 Newton force. To find the strength of the electric field, the 6 Newton force can be divided by the 5 Coloumb charge. To determine the acceleration, Newton's Second Law (F=ma) can be used. The distance between the two charges can be found using the definition of electric force, F = (kq1q2)/d^2, with known values for F, the charges, and the constant k.
  • #1
llama0lover
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A 40 Micro Coluomb test charge is placed in an electric field produced by a 5 Coloumb charge. It expierences a 6 Newton force. What is the strength of the electric field at this point? If the test charge has a mass of 3 x 10 ^ -20 Kg, what acceleration does it expierence?


I Think I could use the 6 Newton force and divide it by the first force 5 Coloumbs to get the electric field, and then on from their I could use a force table get the acceleration, but I am not exactly positive how to set up the Free body diagram and force table for this.

I'm Just stuck on how to find the acceleration really
 
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  • #2
llama0lover said:
I Think I could use the 6 Newton force and divide it by the first force 5 Coloumbs to get the electric field, ...

1. How is force on a charge related to the electric field in which this charge finds itself?
2. 5 coulombs is a charge not a force
3. Assuming that you know the force on the charge and the mass of the charge. What does Newton's Second Law say about the acceleration?
 
  • #3
oh you can use F=ma to figure out the acceleration thanks :-) also, how would you know where the first charge the 5 coluombs was in relation the the test charge?
 
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  • #4
You can find the distance (if you really wanted to) with the definition of electric force, F - (kq1q2)/d2. You know F and the charges, and k's a constant. And distance can't be negative, so you don't have to worry about that when evaluating the square root ;)

(hello classmate! See, I TOLD you I'd get on here!)
 

FAQ: Calculating Electric Field Strength and Acceleration in a Coulombic System

What is an electric field?

An electric field is a physical field produced by electrically charged objects. It describes the influence that a charged object has on other charged objects in its vicinity.

How is an electric field calculated?

The electric field at a specific point in space is calculated by dividing the force exerted on a test charge by the magnitude of the test charge. It is also affected by the distance between the charged objects.

What are some real-life applications of electric fields?

Electric fields have many practical applications, including powering electronic devices, generating electricity, and controlling the movement of charged particles in medical procedures such as MRI scans and radiation therapy.

What is the difference between electric potential and electric field?

Electric potential is the amount of electric potential energy per unit charge at a specific point, while electric field describes the force exerted on a charged object at that point. In other words, electric potential is a scalar quantity while electric field is a vector quantity.

How do electric fields interact with conductors and insulators?

Electric fields interact differently with conductors and insulators. Conductors allow electric charges to move freely, so the electric field inside a conductor is zero. Insulators, on the other hand, do not allow charges to move, so the electric field inside an insulator is non-zero.

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