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Klio2033 [76]
3 years ago
13

Someone please help. Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign a

re placed near each other.
Physics
1 answer:
erik [133]3 years ago
5 0

Answer:

The answer is based on the conservation of energy law; something you should really understand by now.  

For convenience we can hold one of the two charges still; it becomes the frame of reference. And everything we say is in reference to the designated static charge, call it Q.  

So the moving charge, call it q, has total energy TE = PE. It's all potential energy as we start with q not moving.  

It has potential energy because in order to separate q from Q, we had to do work, add energy, on q. And from the COE law, that work added is converted into PE.  

It's a bit like lifting something off the ground. That's work and it becomes GPE. So there's some work, in separating the two charges in the first place.  

But there's more.  

Now we let q go. As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.  

And there's more work, done by the EMF on charge q. That converts the PE into KE and the q charge smashes into Q with some kinetic energy.

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Helpp pls
Kipish [7]

Answer:

The intensity of the electric field is

|E|=10654.37 \:N/C

Explanation:

The electric field equation is given by:

|E|=k\frac{q}{d^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge at 0.4100 m from the balloon
  • d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}

|E|=10654.37 \:N/C

I hope it helps you!

5 0
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