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Net Electrostatic Force on a neg charge next to a pos charge inbetween a field 
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#1
Jan2209, 09:17 PM

P: 3

1. The problem statement, all variables and given/known data
Two point charges, Q1 = 6.9 µC and Q2 = 1.1 µC are located between two oppositely charged parallel plates, as shown in Fig. 1665. The two charges are separated by a distance of x = 0.38 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 72000 N/C. Calculate the net electrostatic force on Q1 and give its direction. 2. Relevant equations Felectric = K (Q1 * Q2) / (r^2) Felectric = qE 3. The attempt at a solution I'm unsure how to approach from this point. I found the force between the two charges. F = (9x10^9) [(6.9 x 10^6)(1.1 x 10^6) / (.38 ^2)] = .4730609418 N Then i figured that because it is a constant electric field I could use this equation. Since the problem states it wants the net charge on Q1, i used that charge for this equation. F = qE = (6.9 x 10^6)(72000) = .4968 N Now I'm trying to find the netforce and whatever I put in seems to be wrong. Electric fields flow from positive to negative so I figured adding these two values would be the correct answer. The force of attraction to Q2 to the right and the force of the electric field to the right. What am I doing wrong? 


#2
Jan2209, 09:48 PM

P: 288

Two forces are acting on Q1. The electric field produced by the plates and the electric field produced by Q2. Consider the signs of the charges. Q1 is negative therefore the direction of the force on Q1 from the plates is to the left and the magnitude of this force is F=QE. The direction of the force on Q1 from Q2 is to the right and it's magnitude is F=KQ1Q2/x^{2}. The force directions are both horizontal so the net force on Q1 can be easily found. I'm assuming the charges are fixed.



#3
Jan2209, 09:52 PM

HW Helper
P: 4,435

In the electric field, the negative charge moves in the opposite direction of the electric field.



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