What is the electric field at the point?

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SUMMARY

The electric field at point A, located at the upper left corner of a square with sides of 0.350 m, is influenced by two equal charges of 7.48 µC positioned at the lower corners. The calculated magnitude of the electric field is 5.49e+5 N/C, derived using the formula kq/r², where k is Coulomb's constant (8.99e9 N m²/C²). The direction of the electric field requires consideration of both charges, and utilizing geometric relationships such as similar triangles can simplify the calculation of force components.

PREREQUISITES
  • Understanding of Coulomb's Law and electric fields
  • Familiarity with the concept of electric field direction
  • Basic knowledge of geometry, specifically similar triangles
  • Proficiency in using the formula kq/r² for electric field calculations
NEXT STEPS
  • Study the concept of vector addition in electric fields
  • Learn how to calculate electric field direction using trigonometric functions
  • Explore the effects of multiple charges on electric fields
  • Review the principles of superposition in electrostatics
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Students studying physics, particularly those focusing on electrostatics, as well as educators seeking to explain electric field calculations involving multiple charges.

Ammora
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Homework Statement



Two tiny objects with equal charges of 7.48 µC are placed at the two lower corners of a square with sides of 0.350 m, as shown.

Image: http://www.webassign.net/grrphys2/16-p-039.gif

Find the electric field at point A, the upper left corner. (Assume the positive x-direction points to the right.)
magnitude: ____N/C
direction: ____° CCW from the positive x-axis

Homework Equations



kq/r^2

The Attempt at a Solution



I already got the magnitude right, here's what I did:
kq/r^2
= 8.99e9 (7.48e-6)/(.350)^2 = 5.49e+5

Where I'm stuck is the direction.
I tried tan theta = (.350) and that was wrong, and I tried tan theta= (.35/2)/.35 and that was wrong also.
Any help would be greatly appreciated. This is due tonight. Thanks in advance.
 
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Ammora said:
I already got the magnitude right, here's what I did:
kq/r^2
= 8.99e9 (7.48e-6)/(.350)^2 = 5.49e+5

That only accounts for the field due to the charge in the lower left corner. What about the other charge?

Also, because you're dealing with a square with sides parallel to the coordinate axes, you should be able to use ratios of side lengths (similar triangles) in lieu of trig functions to work out the force components, if you wish.
 

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