Answered

Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

A point charge gives rise to an electric field with magnitude 1 N/C at a distance of 3 m. If the distance is increased to 6 m, then what will be the new magnitude of the electric field?

Sagot :

onyebuchinnaji

Answer:

The magnitude of the final electric field is 0.5 N/C.

Explanation:

Given;

initial electric field, E₁ = 1 N/C

initial distance moved by the charge, r₁ = 3 m

final distance moved by the charge, r₂ = 6 m

let the magnitude of the final electric field = E₂

The electric potential created by the charge is given as;

V = E₁r₁ = E₂r₂

[tex]E_2 = \frac{E_1r_1}{r_2} \\\\E_2 = \frac{1 N/C\ \times \ 3m}{6m} \\\\E_2 = 0.5 \ N/C[/tex]

Therefore, the magnitude of the final electric field is 0.5 N/C.

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.