What is the repulsive electrical force between two protons in a nucleus that are 5 x 10^ -15 m apart from each other?

Could you help me with this problem? I can't figure out how to do this.

Use Coulomb's equation

F= kQQ/r^2

To calculate the repulsive electrical force between two protons in a nucleus, you can use Coulomb's equation. Let's break it down step by step.

Coulomb's equation states that the force (F) between two charged objects is equal to the product of their charges (Q1 and Q2) multiplied by a constant (k), and divided by the square of the distance between them (r):

F = k * (Q1 * Q2) / r^2

In this case, we are dealing with two protons, which have an identical charge of 1.6 x 10^-19 C (coulombs). The constant (k) is known as the electrostatic constant and has a value of approximately 9 x 10^9 Nm^2/C^2.

Given:
Q1 = 1.6 x 10^-19 C
Q2 = 1.6 x 10^-19 C
r = 5 x 10^-15 m

Now, plug these values into Coulomb's equation:

F = (9 x 10^9 Nm^2/C^2) * ((1.6 x 10^-19 C) * (1.6 x 10^-19 C)) / (5 x 10^-15 m)^2

First, let's calculate the value inside the parenthesis [(1.6 x 10^-19 C) * (1.6 x 10^-19 C)]:

[(1.6 x 10^-19 C) * (1.6 x 10^-19 C)] = 2.56 x 10^-38 C^2

Now substitute this value back into the equation:

F = (9 x 10^9 Nm^2/C^2) * (2.56 x 10^-38 C^2) / (5 x 10^-15 m)^2

Next, square the value of (5 x 10^-15 m) for the denominator:

(5 x 10^-15 m)^2 = (5 x 10^-15 m) * (5 x 10^-15 m) = 25 x 10^-30 m^2 = 2.5 x 10^-29 m^2

Now substitute this value back into the equation:

F = (9 x 10^9 Nm^2/C^2) * (2.56 x 10^-38 C^2) / (2.5 x 10^-29 m^2)

Let's multiply the numbers in the numerator:

(9 x 10^9 Nm^2/C^2) * (2.56 x 10^-38 C^2) = 23.04 x 10^-29 Nm^2/C^2

Finally, divide the numerator by the denominator:

F = (23.04 x 10^-29 Nm^2/C^2) / (2.5 x 10^-29 m^2)

This division can be simplified as follows:

F = 9.216 N

Therefore, the repulsive electrical force between the two protons is approximately 9.216 Newtons.