# Point Charges: direction and magnitude

by matt72lsu
Tags: charges, direction, magnitude, point
 P: 94 1. The problem statement, all variables and given/known data Four point charges are located at the corners of a square with sides of length a. Two of the charges are + q, and two are - q. 1)Find the direction of the net electric force exerted on a charge + Q, located at the center of the square, for the following arrangement of charge: the charges alternate in sign \left( { + q, - q, + q, - q} \right) as you go around the square. 2)Find the magnitude of the net electric force exerted on a charge + Q, located at the center of the square, for the following arrangement of charge: the charges alternate in sign \left( { + q, - q, + q, - q} \right) as you go around the square. 3)Find the magnitude of the net electric force exerted on a charge + Q, located at the center of the square, for the following arrangement of charge: the two positive charges are on the top corners, and the two negative charges are on the bottom corners. Express your answer in terms of the variables q, Q, a, and appropriate constants. 4) Find the direction of the net electric force exerted on a charge + Q, located at the center of the square, for the following arrangement of charge: the two positive charges are on the top corners, and the two negative charges are on the bottom corners. 2. Relevant equations F = k (qQ/r^2) ? 3. The attempt at a solution For question 1 I think it is magnitude = 0 because of the pull directions (b/c the signs alternate). For 4, I think the pull will be downward b/c the Q+ will be attracted to the neg charges. For 2 and 3 I'm not exactly sure what they are asking, so can somebody help me with my comprehension of what they are asking? Thanks!
HW Helper
Thanks
P: 26,148
Hi matt72lsu!
 Quote by matt72lsu For question 1 I think it is magnitude = 0 because of the pull directions (b/c the signs alternate). For 4, I think the pull will be downward b/c the Q+ will be attracted to the neg charges.
Yes, that's right (and that answers 2 also, so that only leaves 3 …)

(but why does the sign alternating make the magnitude 0?)
 For … 3 I'm not exactly sure what they are asking, so can somebody help me with my comprehension of what they are asking?
Find the force on Q from each charge separately.

Force is a vector, so that gives you four vectors … now add them.
 P: 94 Thanks for responding tiny-tim! I think the magnitude is zero because you will have alternating corners "pulling" on the charge, thus it will not "move". I know what I'm trying to say doesn't make any sense... for 2 and 3, i guess I'm just confused on the format of the question. are we actually calculating something or is it just asking me to write an equation for what is happening?
HW Helper
Thanks
P: 26,148
Point Charges: direction and magnitude

 Quote by matt72lsu Thanks for responding tiny-tim! I think the magnitude is zero because you will have alternating corners "pulling" on the charge, thus it will not "move". I know what I'm trying to say doesn't make any sense...
no, it doesn't … if Q was in the centre of just two alternating charges, it would be pulled in the same direction by both, wouldn't it?

Instead, just consider two of the same charges.
 for 2 and 3, i guess I'm just confused on the format of the question. are we actually calculating something or is it just asking me to write an equation for what is happening?
Yes, you are actually calculating something.

You are calculating four vectors, and adding them.
 P: 94 ok i'm sorry i can't quite wrap my head around this one. so am i on the right track with the equation i provided? so r= 0 for 2? for each corner i could label each q1, q2, etc? then find the vector towards Q+ (F1, F2, etc). then add? and k = 8.99 e9 correct
 Sci Advisor HW Helper Thanks P: 26,148 I'm not sure what you're saying , but I think it's correct … show us the actual calculations.
 P: 94 well the directions say to show answer in terma of the variables q, Q, a and k? so i think it would be F= k (qQ/2a^2)

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