Potential due to point charges

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To determine the voltage required to accelerate a proton to penetrate a silicon nucleus, the potential energy is calculated using the formula V = KQ/r, where K is Coulomb's constant, Q is the charge of the silicon nucleus (+14e), and r is the distance from the proton's surface to the nucleus. The effective distance r is considered as the sum of the radii of the proton and the silicon nucleus, approximately 4.8E-15 m. The model answer suggests a potential energy of 4.2E6 J, which aligns with the calculations when interpreting "just penetrate" as the proton's surface barely touching the silicon nucleus. The discussion highlights confusion around the calculations and the interpretation of the parameters involved. Understanding the electric field's dependence on distance from the charge center is crucial for accurate calculations.
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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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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
 
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
 
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.
 
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...
 
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:

\int_S E\cdot dA = \frac{q_{encl}}{\epsilon_0}

For a sphere:

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

E = \frac{q_{encl}}{4\pi \epsilon_0 r^2} = \frac{kq_{encl}}{r^2}

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
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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