How far apart should two protons be if the electrical force of repulsion is equal to the weight (attraction to the Earth) of one of the protons?
Fiqure
is triangle with equal sides 2.5 m and base is 0.5 m
To find the distance between two protons, we need to equate the electrical force of repulsion to the weight (attraction to the Earth) of one of the protons. Let's break down the problem into steps:
Step 1: Calculate the weight of one proton.
The weight of an object can be calculated using the formula:
Weight = mass * gravity
The mass of a proton is approximately 1.67 x 10^-27 kilograms.
The acceleration due to gravity on Earth is roughly 9.8 m/s^2.
Weight of one proton = (1.67 x 10^-27 kg) * (9.8 m/s^2)
Step 2: Calculate the electrical force of repulsion between two protons.
The electrical force between two charged particles can be calculated using Coulomb's Law formula:
Electric Force = (k * q1 * q2) / r^2
In this case, the charges (q1 and q2) are equal for protons, and the value is approximately 1.6 x 10^-19 Coulombs.
The value of the electrostatic constant (k) is approximately 8.99 x 10^9 N m^2 / C^2.
We need to find the distance (r) at which the electrical force of repulsion is equal to the weight.
Step 3: Set up the equation and solve for the distance.
Let the distance between the protons be represented by 'r'.
(8.99 x 10^9 N m^2 / C^2) * (1.6 x 10^-19 C) * (1.6 x 10^-19 C) / r^2 = (1.67 x 10^-27 kg) * (9.8 m/s^2)
Simplifying the equation:
(1.4384 x 10^-28) / r^2 = 1.6376 x 10^-27
Cross multiplying:
1.4384 x 10^-28 = (1.6376 x 10^-27) * r^2
Dividing both sides by (1.6376 x 10^-27):
r^2 = (1.4384 x 10^-28) / (1.6376 x 10^-27)
r^2 ≈ 0.087923192
Taking the square root of both sides:
r ≈ √(0.087923192)
r ≈ 0.2963 meters
Therefore, the distance between the two protons, for which the electrical force of repulsion is equal to the weight of one proton, is approximately 0.2963 meters.