Electrostatic help needed.

[dorjiako]

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Charges 1,2, and 3 are lined up, in that order, at 1mm intervals along the y axis. Three charges are lined up along the axis. Charge 1 has a charge of +4x10^-6C. Charge 2 has a charge of -2x10^-6C. Charge 3 has a charge of -3x10^-6C. What is the change in potential energy of the system if charge 1 is removed.

I tried finding the work done in assembling these charges, multiplied it by -sign to find the potential energy but is at lost at how to remove charge one. I wonder if I am doing this problem wrong. Help needed.

 

[FutureDoctor503]

Do you have the answer to this question? I just want to make sure I don't post anything wrong.

 

[FutureDoctor503]

Is it 126 J? If it is, I'll explain you how I did it.

 

[dorjiako]

How did you do it. Thanks.

 

[FutureDoctor503]

How did you do it. Thanks.

(9*10^9)((-8*10^-12)/0.001)+(6*10^-12)/0.001)+(-12*10^-12)/0.002)) = -72
(9*10^9)*((6*10^-12)/0.001)) = 54

PE = (k*q*Q)/r
PE for 3 charges = (9*10^9)((q1* q2)/distance in meters b/w 1&2)+((q2*q3)/d b/w 2&3)+((q3*q1)/d b/w 1&3) = -72
PE without charge 1 = (9*10^9)*((q2*q3)/d b/w 2&3))= 54
Final PE – initial PE =54-(-72) = 126 J

 

[dorjiako]

Thank you.