Very simply find potential where its zero question

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The discussion revolves around finding points on the axis where the electric potential is zero due to two charges: a +3.0 nC charge at x=0 cm and a -1.0 nC charge at x=4 cm. Initial calculations suggested potential zero at x=3 cm and x=6 cm, but confusion arose regarding calculations in the negative zone. Participants debated whether to use negative distances in their formulas, ultimately confirming that using r1=p and r2=4+p is valid. The final resolution indicated that while p=6 cm was derived, it did not satisfy the equation, whereas p=-6 cm did, leading to further algebraic verification. The conclusion emphasizes the importance of careful algebraic manipulation when determining potential points.
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Very simply "find potential where its zero" question

Homework Statement



A +3.0 nC char age is at x= 0cm and a -1 nC charge is at x= 4cm. At what point or points on the axis is the electric potential zero?

Homework Equations



V= \frac{kQ}{r}

The Attempt at a Solution



i have solved the question, first i tried two places, between the two charges, and after the second charge.

1op3c0.jpg


i solved for x= 3cm and x= 6cm.

now the problem is what is i want to check in the negative zone?

2qxasmp.jpg


now is in
the formula v1+v2 =0

for r1, do i use r1= -p and r2 = 4-p, i am confused about doing it in the negative zone, do i just use the magnitude, if only magnitude, it would be r1=p and r2=4+p, which do i use?
 
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Brown Arrow said:
for r1, do i use r1= -p and r2 = -4-p, i am confused about doing it in the negative zone, do i just use the magnitude, if only magnitude, it would be r1=p and r2=4+p, which do i use?

Other than the underlined [ corrected ] part, both are actually the same. You just multiplied both sides by -1.

So you're fine either way.
 


thrill3rnit3 said:
Other than the underlined [ corrected ] part, both are actually the same. You just multiplied both sides by -1.

So you're fine either way.

i did it, r1=p and r2= 4+p, i get p=6cm, now when i go back and plug in p=6 in the equation it does not satisfy, but =-6cm works, but the answer from solving the equation is p=6cm not p=-6cm, i am confused, help please.0 = \frac{k(3nC)}{P} + \frac{k(-1nC)}{4+P}
 


Brown Arrow said:
i did it, r1=p and r2= 4+p, i get p=6cm, now when i go back and plug in p=6 in the equation it does not satisfy, but =-6cm works, but the answer from solving the equation is p=6cm not p=-6cm, i am confused, help please.0 = \frac{k(3nC)}{P} + \frac{k(-1nC)}{4+P}

I solved for p using the above equation and got p= -6

Maybe something wrong with your algebra?

even with r1= -p and r2= -4-p, I still get p= -6
 
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