Electric Potential at point where Electric Field is Zero

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  • #1
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Homework Statement



a) A -10.1 nC point charge and a +18.9 nC point charge are 13.8 cm apart on the x-axis. What is the electric potential at the point on the x-axis where the electric field is zero?

b) What is the magnitude of the electric field at the two points on the x-axis where the electric potential is zero? (Input your answers in order of increasing distance from the negative point charge

Homework Equations



E = kq/r^2
V=kq/r

where k=9x10^9
r is the radius
q is the charge

The Attempt at a Solution



a) So below is the equation I used but I am getting a negative number for r...someone please help, this is due by this Friday at midnight!

[kq1/(r+d)^2] + [kq2/r^2] = 0
[9x10^9 (-10.1x10-9)/(r + 0.138m)^2] + [9x10^9(18.9x10-9)/r^2] = 0
[-90.9/(r + 0.138m)^2] + [170.1/r^2] = 0
-90.9/(r + 0.138m)^2 = -170.1/r^2
-90.9/-170.1 = (r + 0.138m)^2 / r^2
0.53439 = (r + 0.138m)^2 / r^2
SQRT(0.53439) = (r + 0.138m)/r
0.731202 = r+0.138/r
0.73102r = r+0.138
0.73102r - r = 0.138
-0.268979r = 0.138
r = 0.138/-0.268979
r = -0.51305m

A distance can not be negative, so either my math is wrong or my process of equating the two is wrong. Some help would be greatly appreciated! Then hopefully I will be able to solve part b :)

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Redbelly98
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Homework Statement



a) A -10.1 nC point charge and a +18.9 nC point charge are 13.8 cm apart on the x-axis. What is the electric potential at the point on the x-axis where the electric field is zero?

b) What is the magnitude of the electric field at the two points on the x-axis where the electric potential is zero? (Input your answers in order of increasing distance from the negative point charge

Homework Equations



E = kq/r^2
V=kq/r

where k=9x10^9
r is the radius
q is the charge

The Attempt at a Solution



a) So below is the equation I used but I am getting a negative number for r...someone please help, this is due by this Friday at midnight!
What time zone are you in?
[kq1/(r+d)^2] + [kq2/r^2] = 0
Okay, you do need to consider the possibility that the solution could be at a negative value of the x-coordinate. So r would really be x in this equation.

[9x10^9 (-10.1x10-9)/(r + 0.138m)^2] + [9x10^9(18.9x10-9)/r^2] = 0
Be careful with what is negative and what is positive. E=kq/r2 gives the magnitude of E. You have to use the rule that E points away from a positive charge and towards a negative charge, in order to figure out if the terms are positive or negative.

For example, to the right side of the positive charge, E (due to that charge) points to the right away from the charge, and is negative. But to the left of the positive charge, E (due to that charge) would point to the left to point away from the charge, and would be negative.

[-90.9/(r + 0.138m)^2] + [170.1/r^2] = 0
-90.9/(r + 0.138m)^2 = -170.1/r^2
-90.9/-170.1 = (r + 0.138m)^2 / r^2
0.53439 = (r + 0.138m)^2 / r^2
SQRT(0.53439) = (r + 0.138m)/r
0.731202 = r+0.138/r
0.73102r = r+0.138
0.73102r - r = 0.138
-0.268979r = 0.138
r = 0.138/-0.268979
r = -0.51305m

A distance can not be negative, so either my math is wrong or my process of equating the two is wrong. Some help would be greatly appreciated! Then hopefully I will be able to solve part b :)
 
  • #3
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The only other thing I'll point out is that the square root of something gives a +/- solution.
 
  • #4
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Ok, I am still not understanding what I am doing wrong and why my answer is not right. I have tried with positive and negative values of the square root and still the answer is incorrect. Can someone please help to show me where exactly I am going wrong?
 
  • #5
Redbelly98
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Science Advisor
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Can someone please help to show me where exactly I am going wrong?
Okay.
a) So below is the equation I used but I am getting a negative number for r...someone please help, this is due by this Friday at midnight!

[kq1/(r+d)^2] + [kq2/r^2] = 0
For starters, you seem to think you are looking for "r", the distance from the positive charge. Here is the problem with that thinking: supposed you were to find that r=2 cm. Would that be 2 cm to the right, or 2 cm to the left, of the positive charge?

Instead, realize that we are looking for a point somewhere along the x-axis. So it's better to write the equation in terms of "x", the position along the x-axis, instead of "r".

[9x10^9 (-10.1x10-9)/(r + 0.138m)^2] + [9x10^9(18.9x10-9)/r^2] = 0
Here, you seem to think that E due to the negative charge must always point in the negative direction. And also that E due to the positive charge must point in the positive direction.

Neither of these are necessarily true.
 

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