What is the point where the electric field is zero?

[gnarkil]

Homework Statement

an electron (charge -e) is at the origin, and a particle of charge +5e is at x = 17nm, find a point (x = ...nm) where the electric field is zero.

Homework Equations

e = 1.6*10^-19 coulombs
electric field E = kq/r^2 where k=9*10^9, q is charge, r is distance

The Attempt at a Solution

kq_1/r^2 = kq_2/(17-r)^2

k(-e)/r^2 = k(+5e)/(17-r)^2

k(-e)(17-r)^2 = k(+5e)(r^2)

k(-e)(28 - 34r + r^2) = k(+5e)(r^2)

(9*10^9)(-1.6*10^-19)(28 -34r + r^2) = (9*10^-9)(8*10^-19)(r^2)

-1.4*10^-9[28 - 34r -r^2] = (9*10^-9)(8*10^-19)(r^2)

(-4.032*10^-8) + (4.896*10^-8)r - (1.6*10^-19)r^2 = (7.2*10^-9)r^2

(-4.032*10^-8) + (4.896*10^-8)r - (7.2*10^-9)r^2 = 0

quadratic formula

r = [-4.032*10^-8 +/- sqrt[(4.896*10^-8)^2 - (4)(-7.2*10^-9)(-4.032*10^-8)] ] / (2)(-7.2*10^-9)

r = [-4.89*10^-8 +/- sqrt(1.237*10^-15)] / -1.44*10^-8

r = 0.957nm or 5.83 nm

are my calculations correct? which one is the correct answer?

 

[LowlyPion]

Homework Statement

an electron (charge -e) is at the origin, and a particle of charge +5e is at x = 17nm, find a point (x = ...nm) where the electric field is zero.

Homework Equations

e = 1.6*10^-19 coulombs
electric field E = kq/r^2 where k=9*10^9, q is charge, r is distance

The Attempt at a Solution

kq_1/r^2 = kq_2/(17-r)^2


are my calculations correct? which one is the correct answer?

Perhaps you want to examine your equation a little more carefully.
Are you sure you want to find a point between two charges of opposite sign where the E-field would be 0?

 

[gnarkil]

are you saying the it should be (17+r) instead of (17-r) or asking conceptually? conceptually, if the point is placed a huge distance away, then the electric field would essentially be zero

 

[LowlyPion]

are you saying the it should be (17+r) instead of (17-r) or asking conceptually? conceptually, if the point is placed a huge distance away, then the electric field would essentially be zero

Is there any point between a positive and negative charge that the field can ever be 0?

 

[gnarkil]

no, because the field woulf come out of the positive and flow into the negative charge, correct?

should it be (17+r) not (17-r), though?

 

[gnarkil]

when i made it 17+r the quadratic i would get would be an imaginary number, what am i doing wrong?

 

[LowlyPion]

when i made it 17+r the quadratic i would get would be an imaginary number, what am i doing wrong?

It must be 17+r from the larger charge.

Then you can say e/r² =5e/(17+r)²

Take the sqrt of both sides and rearrange:
(17+r) = (5)^1/2(r)

r = 17/(5^1/2 - 1)

 

[gnarkil]

oh, that simple i got 13.75, but actually it was -13.75

 

[LowlyPion]

oh, that simple i got 13.75, but actually it was -13.75

You found r, but that is the distance along the -x direction. So yes the position would be -r in that direction.