Electric Field Strength Between Two Charges

  • Thread starter rnjscksdyd
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  • #1

Homework Statement


Two isolated point charges, -7 μC and +2 μC, are at a fixed distance apart. At which point is it possible for the electric field strength to be zero?

(A)............................(B)............(C)..................(D)
o----------------------o----------o---------------o
................................-7 μC.......+2 μC

I don't know how to insert image
sorry for messy drawing (I had to put little dots because it wouldn't make spaces although I spaced them out)

Homework Equations


F= k(q1)(q2)/r^2
E = F/q

The Attempt at a Solution


The actual answer is (D).

By these two equations, I got
E = kq/r^2

I thought, to make the field strength 0, I had to find the distance at which the field strength of these two charges are equal.

∴ E = E
kq/r^2 = kq/r^2
k(-7 μC )/r^2 = k(+2 μC)/r^2
I know this is way off,,, since I will cancel out the distance!!
I actually did choose the right answer (D) simply because -7 μC has larger magnitude that field strength would be much smaller at (D)

Please tell me how I can solve this algebraically. Thank you :D
 

Answers and Replies

  • #2
tiny-tim
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hi rnjscksdyd! welcome to pf! :smile:
I thought, to make the field strength 0, I had to find the distance at which the field strength of these two charges are equal.
correct :smile:
∴ E = E
kq/r^2 = kq/r^2
k(-7 μC )/r^2 = k(+2 μC)/r^2
no, it isn't the same r, is it? :wink:
 
  • #3
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Try and us the Principle of Superposition. That might lead you to the answer, pal! :)
 
  • #4
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Let d represent the distance between the two charges, and let x represent the distance to the right of the 2 μC charge at which the field is zero. In terms of x and d, how for is the -7 μC charge from the point where the field is zero? You will be solving for x in terms of d.
 
  • #5
hi rnjscksdyd! welcome to pf! :smile:


correct :smile:


no, it isn't the same r, is it? :wink:
I thought 'r' was a distance between the two charges. So, 'r' should be different in that, say, 'r1' is the distance between -7μC and the point where electric field strength is 0 and 'r2' is the distance between +2μC and the point where electric field strength is 0, right??

Try and us the Principle of Superposition. That might lead you to the answer, pal! :)
Could you explain further please :( I thought superposition was for the waves, like constructive and destructive interference :( how do i apply this concept in electric field?

Let d represent the distance between the two charges, and let x represent the distance to the right of the 2 μC charge at which the field is zero. In terms of x and d, how for is the -7 μC charge from the point where the field is zero? You will be solving for x in terms of d.
would that mean:
-7μC/(d+x) = +2μC/(x) ?

If it is so, how do I know that the answer is D? From the above equation I got:
-7μC/(d+x) = +2μC/(x)
-7μCx = +2μC(d+x)
x = -2d/9

Also, how do I know that 'x' is the distance to the right of +2μC supposing I did not know the answer?

Sorry I'm just bad at physics :(
 
  • #6
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4,396
would that mean:
-7μC/(d+x) = +2μC/(x) ?
That's not the equation I get. If I do a force balance on a test charge, I get:

2μC/(x)2=+7μC/(d+x)2

Note the squares in the denominators. Next, take the square root of both sides of the equation.

Also, how do I know that 'x' is the distance to the right of +2μC supposing I did not know the answer?

Sorry I'm just bad at physics :(
It doesn't matter. The algebra would take care of that anyway.

Chet
 
  • #7
hi rnjscksdyd! welcome to pf! :smile:


correct :smile:


no, it isn't the same r, is it? :wink:
Try and us the Principle of Superposition. That might lead you to the answer, pal! :)
Let d represent the distance between the two charges, and let x represent the distance to the right of the 2 μC charge at which the field is zero. In terms of x and d, how for is the -7 μC charge from the point where the field is zero? You will be solving for x in terms of d.
That's not the equation I get. If I do a force balance on a test charge, I get:

2μC/(x)2=+7μC/(d+x)2

Note the squares in the denominators. Next, take the square root of both sides of the equation.



It doesn't matter. The algebra would take care of that anyway.

Chet
Sorry, forgot about the squares :D
Then, is this right?
[tex]\frac{-7μC}{(d+x)^2}= \frac{+2μC}{x^2}[/tex]
[tex]-7μCx^2 = +2μC(d+x)^2[/tex]
[tex]-7μCx^2 = +2μC(d^2 + 2dx + x^2)[/tex]
[tex]-7x^2 = 2d^2 + 4dx + 2x^2[/tex]
I took out μC for now (it looks better :D)
[tex]0 = 9x^2 + 4dx + 2d^2[/tex]

[itex]x=\frac{-4d \pm \sqrt{16d^2 - 72d^2}}{18}[/itex]

but then I get a complex number since I get negative square root. Where did I go wrong?
 
  • #8
tiny-tim
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hi rnjscksdyd! :smile:

(just got up :zzz:)
I took out μC for now (it looks better :D)
it does! … and you could have taken it out from the start :wink:
[itex]x=\frac{-4d \pm \sqrt{16d^2 - 72d^2}}{18}[/itex]

but then I get a complex number since I get negative square root. Where did I go wrong?
your work (and result) is perfect except

can you think of a situation in which it isn't (d + x) ? :smile:
 
  • #9
20,523
4,396
Sorry, forgot about the squares :D
Then, is this right?
[tex]\frac{-7μC}{(d+x)^2}= \frac{+2μC}{x^2}[/tex]
[tex]-7μCx^2 = +2μC(d+x)^2[/tex]
[tex]-7μCx^2 = +2μC(d^2 + 2dx + x^2)[/tex]
[tex]-7x^2 = 2d^2 + 4dx + 2x^2[/tex]
I took out μC for now (it looks better :D)
[tex]0 = 9x^2 + 4dx + 2d^2[/tex]

[itex]x=\frac{-4d \pm \sqrt{16d^2 - 72d^2}}{18}[/itex]

but then I get a complex number since I get negative square root. Where did I go wrong?
You still have that minus sign in front of the 7, which is incorrect. It should be a + sign.
[tex]\frac{7μC}{(d+x)^2}= \frac{+2μC}{x^2}[/tex]
Taking the square root of both sides:
[tex]\frac{(d+x)}{x}=\sqrt{3.5}[/tex]
 
  • #10
Sorry for late reply, my internet was down -_-;; this country :(
hi rnjscksdyd! :smile:

(just got up :zzz:)


it does! … and you could have taken it out from the start :wink:


your work (and result) is perfect except

can you think of a situation in which it isn't (d + x) ? :smile:
:( what do you mean when it isn't (d+x)?

You still have that minus sign in front of the 7, which is incorrect. It should be a + sign.
[tex]\frac{7μC}{(d+x)^2}= \frac{+2μC}{x^2}[/tex]
Taking the square root of both sides:
[tex]\frac{(d+x)}{x}=\sqrt{3.5}[/tex]
Does that mean... when that value is negative, the field will be 0 on the left side and on the right if the value is positive?
 
  • #11
tiny-tim
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hi rnjscksdyd! :smile:
:( what do you mean when it isn't (d+x)?
if it's d+x on the left, isn't it d-x on the right? :wink:
 
  • #12
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Could you explain further please :( I thought superposition was for the waves, like constructive and destructive interference :( how do i apply this concept in electric field?
'According to this principle the total amount of force on a given particle is the vector sum of the individual forces applied on that charged particle by all the other charged particles. The individual forces of the particles is calculated using the formula of Coulombs law and is not influenced by the presence of other particles.'

Basically, Find the Coulomb's Force acting on each particle, one at a time, and then add it vectorially.
(Rememeber, DO NOT FORGET THE SIGN OF THE CHARGE OF THE PARTICLE!)
 

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