Potential due to point charges

In summary, the question asks for the voltage needed to accelerate a proton so that it can just penetrate a silicon nucleus. The answer is 4.2E6 J and it is found by using the equation V = KQ/r, where r is the sum of the proton and silicon nucleus radii. This is because the electric field between charges is determined by the distance from their centres, and both the proton and silicon nucleus are assumed to be uniform spheres.
  • #1
crazy student
7
0
How much voltage must be used to accerlerate a proton (radius 1.2E-15 m) so that it has sufficient energy to just penetrate a silicon nucleus? A silicon nucleus has a charge of +14e, and its radius is about 3.6E-15 m. Assume the potential is that for point charges.
 
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  • #2
crazy student said:
How much voltage must be used to accerlerate a proton (radius 1.2E-15 m) so that it has sufficient energy to just penetrate a silicon nucleus? A silicon nucleus has a charge of +14e, and its radius is about 3.6E-15 m. Assume the potential is that for point charges.
What is the potential energy of a charge q1 a distance r from another charge q2? What is the potential energy of a proton whose surface is situated 3.6E-15 m. from the centre of a silicon nucleus?

AM
 
  • #3
that's all from the question given...
I think it suppose r starts from infinity
but i can't get to the model ans 4.2E-6 J
 
  • #4
crazy student said:
that's all from the question given...
I think it suppose r starts from infinity
but i can't get to the model ans 4.2E-6 J

You need to check the exponent in that answer, and you need to interpret "just penetrate" as the outer surface of the proton just barely penetrating the outer surface of the silicon nucleus.
 
  • #5
sorry, it should be 4.2E6 J
I found the question quite confusing also...
from the model ans and equation V = KQ/r
r = 4.8E-15 m, which is the sum of proton and silicon nucleus radii
but i don't know the reason behind...
 
  • #6
crazy student said:
sorry, it should be 4.2E6 J
I found the question quite confusing also...
from the model ans and equation V = KQ/r
r = 4.8E-15 m, which is the sum of proton and silicon nucleus radii
but i don't know the reason behind...
The force between charges depends on the distance from the centre of the charge. The electric field of a charge is given by Gauss' law, which states that the field depends only on the enclosed charge:

[tex]\int_S E\cdot dA = \frac{q_{encl}}{\epsilon_0}[/tex]

For a sphere:

[tex]\int_S E\cdot dA = 4\pi r^2E = \frac{q_{encl}}{\epsilon_0}[/tex]

[tex]E = \frac{q_{encl}}{4\pi \epsilon_0 r^2} = \frac{kq_{encl}}{r^2}[/tex]

So if you assume that a proton and a silicon nucleus are uniform spheres, the field at their surface depends only upon the distance of that surface from the respective centres.

AM
 

1. What is potential due to point charges?

Potential due to point charges is the measure of the electrical potential energy of a point charge in an electric field. It is the amount of work required to move a unit positive charge from infinity to a specific point in the electric field.

2. How is potential due to point charges calculated?

The potential due to point charges can be calculated using the formula V = kQ/r, where V is the potential, k is the Coulomb constant, Q is the magnitude of the point charge, and r is the distance from the point charge.

3. What is the unit of potential due to point charges?

The unit of potential due to point charges is volts (V). It is a derived unit in the International System of Units (SI) and is equivalent to joules per coulomb (J/C).

4. How does the distance from a point charge affect potential due to point charges?

The potential due to point charges decreases as the distance from the point charge increases. This is because the electric field strength decreases with distance, resulting in a smaller potential at a farther distance.

5. Can the potential due to point charges be negative?

Yes, the potential due to point charges can be negative. This occurs when the point charge has a negative electric charge, which results in an attractive force towards the point charge. The potential is negative because work must be done to move a positive charge towards the negative charge against the direction of the electric field.

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