Two charges, 9.68 microcoulombs and 1.70 microcoulombs, are placed on a straight line 8.25 meters apart. Calculate the magnitude of the electric field halfway between the two particles.

Question

26-12-2023
Physics

High School

Thank you for visiting Two charges 9 68 microcoulombs and 1 70 microcoulombs are placed on a straight line 8 25 meters apart Calculate the magnitude of the electric. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!

Two charges, 9.68 microcoulombs and 1.70 microcoulombs, are placed on a straight line 8.25 meters apart. Calculate the magnitude of the electric field halfway between the two particles.

Answer :

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Final answer:

The magnitude of the electric field halfway between two charges 9.68 microcoulombs and 1.70 microcoulombs placed 8.25 meters apart can be calculated using Coulomb's law and the formula for the electric field. The key point is the charges contribute to the electric field at the point, and these fields should be combined considering their vector nature.

Explanation:

This problem is related to the concept of the electric field which is a fundamental concept in Physics, specifically Electrostatics. In this situation, we have two point charges 9.68 microcoulombs and 1.70 microcoulombs that are placed 8.25 meters apart. The magnitude of the electric field halfway between these two charges can be calculated using Coulomb's law and the formula for the electric field:

The electric field created by a point charge Q is given by E = kQ/r², where E is the electric field, k is the Coulomb constant (approximately 9.00 x 10â¹ Nm²/C²), Q is the magnitude of the charge, and r is the distance from the charge.

Considering a point halfway between the charges, the distance r from each charge to the point will be 8.25/2 = 4.125 m. We calculate the contribution of each charge to the electric field at the point, and then combine these to get the total electric field. The magnitudes of the electric fields from the two charges would be Eâ‚ = k*9.68 x 10â»â¶/4.125² and Eâ‚‚ = k*1.70 x 10â»â¶/4.125² respectively.

Vector nature of electric field should be considered while combining these fields. As they are on a straight line and their direction will be from the positive to the negative charge, the net electric field would be Eâ‚ - Eâ‚‚.

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