(a) What is the electric potential a distance of 2.3 10-15 m away from a proton?

Answer for this part is :626000

(b) What is the electric potential energy of a system that consists of two protons 2.3 10-15 m apart-as might occur inside a typical nucleus?

Answer this part please
i just need the final answer

Oh, I see you're exploring the electrifying world of protons and their potential! Now, when two protons get cozy inside a typical nucleus, their electric potential energy can be quite electrifying. So, without further ado, the final answer to the electric potential energy of this system at a distance of 2.3 x 10^-15 m apart is... drum roll, please... 1.25 x 10^-6 Joules! May the protons continue to stay positively charged in their nuclear dance! 🎉

To find the electric potential energy of a system that consists of two protons, we can use the formula:

Electric potential energy = (k * q1 * q2) / r

Where:
- k is the electrostatic constant, approximately 9.0 x 10^9 Nm^2/C^2
- q1 and q2 are the charges of the protons, which are both equal to the elementary charge, approximately 1.6 x 10^-19 C
- r is the distance between the protons, which is given as 2.3 x 10^-15 m

Substituting the values into the formula, we have:

Electric potential energy = (9.0 x 10^9 Nm^2/C^2) * (1.6 x 10^-19 C)^2 / (2.3 x 10^-15 m)

Calculating this expression gives us approximately -2.18 x 10^-12 Joules.

So, the final answer for the electric potential energy of the system is approximately -2.18 x 10^-12 Joules.

To calculate the electric potential energy of a system consisting of two protons, we can use the formula:

Electric potential energy (U) = (k * q1 * q2) / r

Where:
k is the Coulomb's constant (9 x 10^9 N m^2/C^2),
q1 and q2 are the charges of the protons (1.6 x 10^-19 C),
and r is the distance between the protons (2.3 x 10^-15 m).

Plugging in the values, we get:

U = (9 x 10^9 N m^2/C^2) * (1.6 x 10^-19 C) * (1.6 x 10^-19 C) / (2.3 x 10^-15 m)

To get the final answer, we can calculate this expression.

Vq= PE

kq1*q2/r=PE