Electric Potential/Potential Energy problem

In summary, the first problem involves a particle with mass 6 grams and charge 54 μc moving at 25 m/s towards a stationary particle with mass 13 grams and charge 22 μc. The question asks for the distance of closest approach between the two particles. The second problem involves four identical particles with charge 4 μC and mass 2 grams, released from rest at the vertices of a square with side length 42 cm. The question asks for the speed of each particle when their distance from the center of the square is tripled. Both problems can be solved using the concept of conservation of momentum and energy.
  • #1
jawhnay
37
0

Homework Statement



a) From a large distance away, a particle of mass 6 grams and charge 54 μc is fired at 25 m/s to the right straight toward a second particle, originally stationary but free to move, with mass 13 grams and charge 22 μc. Find the distance of closest approach between the charges.

b) Four identical particles each have charge of 4 μC and mass 2 grams. They are released from rest at the vertices of a square of side 42 cm. How fast is each particle moving when their distance from the center of the square triples?

Homework Equations


a) ΔU + ΔK = 0
b) U=(kQ1Q2)/r , ΔU + ΔK = 0, K = (1/2)mv2, V = (KQ)/r

The Attempt at a Solution


a) not sure how to do this problem other than knowing that Vi = 25 m/s for the 6 gram particle and Vi = 0 for the 13 gram particle.
 
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  • #2
jawhnay said:

Homework Statement



a) From a large distance away, a particle of mass 6 grams and charge 54 μc is fired at 25 m/s to the right straight toward a second particle, originally stationary but free to move, with mass 13 grams and charge 22 μc. Find the distance of closest approach between the charges.

b) Four identical particles each have charge of 4 μC and mass 2 grams. They are released from rest at the vertices of a square of side 42 cm. How fast is each particle moving when their distance from the center of the square triples?


Homework Equations


a) ΔU + ΔK = 0
b) U=(kQ1Q2)/r , ΔU + ΔK = 0, K = (1/2)mv2, V = (KQ)/r


The Attempt at a Solution


a) not sure how to do this problem other than knowing that Vi = 25 m/s for the 6 gram particle and Vi = 0 for the 13 gram particle.
For (a), how can you describe the energy at their closest approach?

(b) can be solved in a similar way.
 
  • #3
I think (a) also requires conservation of momentum.
 
  • #4
mfb said:
For (a), how can you describe the energy at their closest approach?

(b) can be solved in a similar way.

i'm not sure... my knowledge of energy is kind of rusty since i haven't taken mechanics in a while...
 
  • #5
voko said:
I think (a) also requires conservation of momentum.
Right
jawhnay said:
i'm not sure... my knowledge of energy is kind of rusty since i haven't taken mechanics in a while...
Then you should try to fix that. While it is possible to solve it without the concept of energy, this would require solving differential equations, which is NOT easier.
 

1. What is electric potential?

Electric potential is the amount of electric potential energy per unit charge at a specific point in an electric field. It is measured in volts (V) and represents the amount of work required to move a unit charge from one point to another in the electric field.

2. How is electric potential different from electric potential energy?

Electric potential energy is a measure of the potential energy that an object with a charge has in an electric field. It is a scalar quantity, while electric potential is a vector quantity. Electric potential energy is dependent on both the magnitude of the charge and its position in the electric field, while electric potential is only dependent on the position.

3. What is the formula for calculating electric potential?

The formula for electric potential is V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge, and r is the distance between the charge and the point where the potential is being measured. This formula assumes a point charge and can be used to calculate the potential at any point in an electric field.

4. Can electric potential be negative?

Yes, electric potential can be negative. This means that the electric field is attractive and the direction of the electric force is towards the point where the potential is being measured. A positive electric potential indicates a repulsive electric field.

5. How is electric potential related to electric potential energy?

Electric potential and electric potential energy are closely related. Electric potential energy is equal to the product of electric potential and charge, so an increase in electric potential will result in an increase in electric potential energy. Additionally, the change in electric potential energy between two points is equal to the difference in electric potential between those points multiplied by the charge.

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