What is the magnitude of the electric field?

AI Thread Summary
To determine the magnitude of the electric field 10 cm to the right of the negative charge, both the positive and negative charges must be considered due to the superposition principle. The electric field from the positive charge (+25 µC) and the negative charge (-11 µC) must be calculated separately and then combined. The incorrect answer may stem from not accounting for the direction of the electric fields produced by each charge. The correct approach involves calculating the electric field contributions from both charges at the specified point and summing them vectorially. Understanding these principles is crucial for solving the problem accurately.
zoomba91210
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Homework Statement


Two charged particles are fixed on the x-axis 11 cm apart. The one on the left has a net charge of +25uc and the one on the right has a net charge of -11uc. What is the magnitude of the electric field 10 cm to the right of the negative charge?

Homework Equations


E = (kq)/r^2

The Attempt at a Solution


When I plug in 9.0 x 10^9 for k, 11.0 x 10^-6 C for q and 0.1 m for r, I get the wrong answer.

Thanks for any help!
 
Last edited:
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Hello, zoomba91210.

Both charges contribute to the electric field at the point located 10 cm to the right of the negative charge. ("Superposition principle")
 
Last edited:
Suppose that a +ve unit test charge is located at a point on which E is to be calculated. Find the net force on that charge due to the two charges.
 

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