Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

The nucleus in an iron atom has a radius of about 4.0 10-15 m and contains 26 protons. (a) What is the magnitude of the repulsive electrostatic force between two of the protons that are separated by 4.0 10-15 m?

Sagot :

onyebuchinnaji

Answer:

The  repulsive electrostatic force between two of the protons is 14.4 N.

Explanation:

Given;

distance between the two protons, r = 4 x 10⁻¹⁵ m

charge of a proton, q = + 1.6 x 10⁻¹⁹ C

The repulsive electrostatic force between two of the protons is calculated by applying Coulomb's law;

[tex]F = \frac{k q_1 q_2 }{r^2} \\\\where;\\k \ is \ Coulomb's \ constant = 9\times 10^9 \ Nm^2/C^2 \\\\F = \frac{9\times 10^9 \ (1.6\times 10^{-19})^2}{(4 \times 10^{-15})^2} \\\\F = 14.4 \ N[/tex]

Therefore, the  repulsive electrostatic force between two of the protons is 14.4 N.

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.