Electric potential energy and particle charge

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  • #1
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I've tried all three of these problems I don't know how mant times, and my work is just not correct. Any clue? The third prob, i really used some awkward equations.

A particle has a charge of +1.53 mC and moves from point A to point B, a distance of 0.197 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA-EPEB=+9.39×10-4 J. Find (a) the magnitude of the electric force that acts on the particle and (b) the magnitude of the electric field that the particle experiences.

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.0680 V exists across the membrane. The thickness of the membrane is 7.95×10-9 m. What is the magnitude of the electric field in the membrane?

Two hollow metal spheres are concentric with each other. The inner sphere has a radius of 0.1440 m and a potential of 86.0 V. The radius of the outer sphere is 0.146 m and its potential is 82.0 V. If the region between the spheres is filled with Teflon, find the electric energy contained in this space.
 
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  • #2
1) A particle has a charge of +1.53 mC and moves from point A to point B, a distance of 0.197 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA-EPEB=+9.39×10-4 J. Find (a) the magnitude of the electric force that acts on the particle and (b) the magnitude of the electric field that the particle experiences.

i derived force by solving for E=kq/rsquared, then plugged into F=qE.
i derived electric field by solving for the change in V=change in potentials/q, then i found teh electric field=-change in V/change in distance.

I'm guessing i used the wrong equations somewhere b/c I am wrong in my process.


2) The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.0680 V exists across the membrane. The thickness of the membrane is 7.95×10-9 m. What is the magnitude of the electric field in the membrane?

I thought i could just use E=-change in V/change in distance, but my solution was incorrect.

3) Two hollow metal spheres are concentric with each other. The inner sphere has a radius of 0.1440 m and a potential of 86.0 V. The radius of the outer sphere is 0.146 m and its potential is 82.0 V. If the region between the spheres is filled with Teflon, find the electric energy contained in this space.

This one i played with, and someone gave me an equation to work with but, what this is equal to and where they derived this i have no clue
[(q/4piEo)(outer radius-inner radius)]/(outer radius*inner radius)

so i dint bother to deal with this equation, even though it looks all simplified and easy.
b/c i thought i could find a solution using V=(kq/rsquared)+(k/q'/rsquared).
but that's not the easiest direction.
 
  • #3
figs said:
1) A particle has a charge of +1.53 mC and moves from point A to point B, a distance of 0.197 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA-EPEB=+9.39×10-4 J. Find (a) the magnitude of the electric force that acts on the particle and (b) the magnitude of the electric field that the particle experiences.

i derived force by solving for E=kq/rsquared, then plugged into F=qE.

You do not need to determine the force field of the moving charge, but the electric field that acts upon it. Just think what is r you used in your formula.

Read the text: the particle experiences a constant electric force.
Now you have to know how to calculate the the difference of the potential energy between points A and B. It is equal the work done by the electric force. It is constant and parallel to the line of motion, so the work is simple force times distance. You know work and distance, you get the force.

You can determine the magnitude of the electric field now. The electric field intensity is defined as the electric force on a charged particle divided by its charge. You know the force and you know the charge. So you can get the field intensity.

ehild
 

Related to Electric potential energy and particle charge

What is electric potential energy?

Electric potential energy is the potential energy that a charged particle has due to its position in an electric field. It is the ability of a charged particle to do work.

How is electric potential energy calculated?

Electric potential energy is calculated using the formula U = qV, where U is the electric potential energy, q is the charge of the particle, and V is the electric potential. The unit for electric potential energy is joules (J).

What is the relationship between electric potential energy and particle charge?

The electric potential energy of a charged particle is directly proportional to its charge. This means that as the charge of the particle increases, its electric potential energy also increases.

How does the distance between charged particles affect electric potential energy?

The electric potential energy between two charged particles decreases as the distance between them increases. This is because the electric force between them weakens as the distance increases.

How is electric potential energy used in practical applications?

Electric potential energy is used in many practical applications, such as in batteries, which convert chemical energy into electric potential energy to power devices. It is also used in electric motors, generators, and other electrical devices.

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