Electric field strength at a point due to 3 charges

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The discussion revolves around calculating the electric field strength at a point due to three charges, resulting in a proposed value of 13q for the new charge needed to balance the electric field to zero. The user applied the electric field equation and equated the contributions from the existing charges, seeking verification of their calculations. Other participants suggest providing more detailed steps in the solution to clarify the reasoning, particularly regarding the absence of 'x' in the equation and the sign choices for the terms. Overall, the consensus leans towards the correctness of the user's answer, pending further elaboration on the methodology. The conversation emphasizes the importance of clear problem-solving steps in physics calculations.
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Homework Statement
Find the value of the charge placed at the point that makes the electric field 0.
Relevant Equations
Electric field: Kq /r^2
I got E. 13q as the answer. That is what i did: The electric field due to +q at origin 0 should equal the electric fields of charges -3q and the new charge placed at 2x. So applying the equation above like this; k*(q) / (2^2) = -3q*k + (k*C)/ 4 solving for C the new charge added, gives 13q. I don't know if it's correct because i don't have the answer to this question . Can you please check my working thanks.
 

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Tesla In Person said:
Homework Statement:: Find the value of the charge placed at the point that makes the electric field 0.
Relevant Equations:: Electric field: Kq /r^2

I got E. 13q as the answer. That is what i did: The electric field due to +q at origin 0 should equal the electric fields of charges -3q and the new charge placed at 2x. So applying the equation above like this; k*(q) / (2^2) = -3q*k + (k*C)/ 4 solving for C the new charge added, gives 13q. I don't know if it's correct because i don't have the answer to this question . Can you please check my working thanks.
I think your answer is correct.
 
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In order to make it easier for someone to follow your solution, you could show more steps. In particular, show why x does not appear in your equation and explain the choice of signs for the terms in your equation.
 
TSny said:
I think your answer is correct.
It must be if it really is "Tesla".
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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