What Is the Magnitude and Sign of Earth's Electric Charge?

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Homework Help Overview

The discussion revolves around determining the magnitude and sign of Earth's electric charge based on the electric field measurements at its surface. The subject area includes concepts from electrostatics and electric fields.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between electric fields and charge, questioning how to find the magnitude of Earth's charge given the electric field strength. Some discuss the implications of field line direction in relation to charge sign.

Discussion Status

Participants are actively engaging with the problem, with some offering guidance on using the electric field definition related to point charges. There is a recognition of the need for additional information to fully resolve the question.

Contextual Notes

Some participants note a lack of sufficient information to determine the charge magnitude, and there is an emphasis on understanding the assumptions related to the uniform charge distribution of the Earth.

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I am having trouble doing this question.

Measurements indicate there is an electric field surrounding the earth. Its magnitude is about 100 N/C at the Earth's surface and points to the centre of the earth. What is the magnitude of the electric charge on the earth? Is it positive or negative?
(Hint: The electric field caused by a uniformly charged sphere is the same as if the entire charge were concentrated at its centre)
 
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Look in your book and find out which direction field lines point. Do they leave positive charges and end on negative charges? Or vice versa?

- Warren
 
but how do I find the magnitude of the electric charge on earth?
I don't have enought information
 
Use the definition of the field due to a point charge. Pretend that all of the Earth's charge is concentrated at a point at the center of the earth. The Earth's radius is 6,378.1 kilometers.

[tex]E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}[/tex]

- Warren
 

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