Lectures on Physics has been derived from Benjamin Crowell's Light and Matter series of free introductory textbooks on physics. See the editorial for more information.... 
Home Electricity Fields of Force Examples Superposition of electric fields  
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Superposition of electric fields
Question: Charges q and q are at a distance b from each other, as shown in the figure. What is the electric field at the point P, which lies at a third corner of the square? Solution: The field at P is the vector sum of the fields that would have been created by the two charges independently. Let positive x be to the right and let positive y be up. Negative charges have fields that point at them, so the charge q makes a field that points to the right, i.e. has a positive x component. Using the answer to the selfcheck, we have E _{q,x} = kq/b^{2} E _{q,y} = 0 . Note that if we had blindly ignored the absolute value signs and plugged in q to the equation, we would have incorrectly concluded that the field went to the left. By the Pythagorean theorem, the positive charge is at a distance √2b from P, so the magnitude of its contribution to the field is kq/(√2b) ^{2} = kq/2b^{2}. Positive charges have fields that point away from them, so the field vector is at an angle of 135° counterclockwise from the x axis.
The total field is


Home Electricity Fields of Force Examples Superposition of electric fields 