Where is the electric field zero?

[CatWhisperer]

Homework Statement

Two point charges are placed 1.00m apart.
q(1) = -2.50 x 10^(-6) C
q(2) = +6.00 x 10^(-6) C
Task is to find where along the line, other than at infinity, the electric field will be equal to zero.

Homework Equations

E = (k * q) / r^2

The Attempt at a Solution

I let E(1) + E(2) = 0 and substituted the above formula in, which, after simplifying gives me a quadratic (I let 'x' equal the first radius, and '1.00 - x' equal the second, so q(1) is at zero on the axis, and the solution for x will give me the point on the axis with respect to q1 at which the two fields cancel out and the net field equals zero).

I solved the quadratic and came up with:

x = 0.39m (0.39m to the right of q(1) ) and
x = -1.83m (1.83m to the left of q (1) )

The answer is -1.83m, but I am wondering if someone can explain why +0.39m isn't correct also? I think my conceptual understanding is lacking here, and my textbook hasn't quite revealed the answer.

Thanks very much in advance, folks.

 

[SammyS]

Homework Statement

Two point charges are placed 1.00m apart.
q(1) = -2.50 x 10^(-6) C
q(2) = +6.00 x 10^(-6) C
Task is to find where along the line, other than at infinity, the electric field will be equal to zero.

Homework Equations

E = (k * q) / r^2

The Attempt at a Solution

I let E(1) + E(2) = 0 and substituted the above formula in, which, after simplifying gives me a quadratic (I let 'x' equal the first radius, and '1.00 - x' equal the second, so q(1) is at zero on the axis, and the solution for x will give me the point on the axis with respect to q1 at which the two fields cancel out and the net field equals zero).

I solved the quadratic and came up with:

x = 0.39m (0.39m to the right of q(1) ) and
x = -1.83m (1.83m to the left of q (1) )

The answer is -1.83m, but I am wondering if someone can explain why +0.39m isn't correct also? I think my conceptual understanding is lacking here, and my textbook hasn't quite revealed the answer.

Thanks very much in advance, folks.

What specifically are you using for E(1) & E(2) ?

 

[CatWhisperer]

I have substituted the formula E = k(e) * q / r^2

So I get:

E(net) = E(1) + E(2)

Where

E(1) = k(e) * q(1) / r(1)^2
E(2) = k(e) * q(2) / r(2)^2

k(e) = 8.99 * 10^9
r(1) = x
r(2) = 1.00 - x
q(1) & q(2) as given in the OP

 

[SammyS]

I have substituted the formula E = k(e) * q / r^2

So I get:

E(net) = E(1) + E(2)

Where

E(1) = k(e) * q(1) / r(1)^2
E(2) = k(e) * q(2) / r(2)^2

k(e) = 8.99 * 10^9
r(1) = x
r(2) = 1.00 - x
q(1) & q(2) as given in the OP

So, you have q(1) at the origin, and q(2) on the x-axis at x = 1 meter.

You also have E(1) being negative and E(2) being positive, (but all the algebra "cares about" is that they have opposite sign). Those signs will be that way only to the right of q(2). They wil both be opposite that to the left of q(1) so the results will still be good there.

Between q(1) and q(2), both E(1) and E(2) point to the left, i.e., they're both negative.

 

[CatWhisperer]

Gotcha. The fields can't cancel out unless they point in opposite directions, which happens on the left of q1 (and the right of q2, but then other conditions are not met), but not between q1 and q2 :-)

Thank you!