Electric Field at Point B: q, L & Q Effects

AI Thread Summary
The discussion focuses on calculating the electric field at point B, located halfway between two opposite charges in a square configuration. The total electric field is derived as E = (36/5)(q/L^2), taking into account the contributions from all four charges. For the acceleration comparison, both objects experience the same electric force per unit mass, but the object with charge Q and mass M/2 will accelerate faster due to its lower mass. The electric field lines are suggested to be drawn based on the interactions of the two positive and two negative charges, reflecting their opposing effects. The discussion also touches on potential equilibrium points, indicating that the electric field is not zero at point B, and further analysis is needed to determine the stability of any equilibrium points.
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Homework Statement


Four charges, each q in magnitude, are arranged in a square, two positive and two negative, in opposite corners. The square is L length on each side, and the charges are all in the same plane as shown in the diagram. Show work and units for full credit.

q+ -----B----- q-
| |
| | L
| |
q- ------------ q+
L
**vertical dotted line makes square but spacing corrects itself

A. Determine the magnitude and direction of the total electric field at point B (half way between the top two charges) due to all four charges as a function of q and L.

B. If placed at point B, which would accelerate faster, object 1 with a charge 2Q and mass M or object 2 with a charge Q and mass M/2? Why?

C. Draw the electric field lines in the plane of the square indicating the direction and general shape of the electric field generated by this charge arrangement. Hint: think about the electric field generated by two opposite charges.

D. Is the electric field zero at any point(s)? If so, where? Also, indicate any equilibrium point(s) and whether they are stable, unstable or neutral.
 
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A. We have made it to E=(36/5)(q/L^2)
B. Both are equal a = 2qE/m = qE/(m/2)
 
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