# Two charges +1*10^-6 and -4*10^-6 are separatd by a distance of 2 m. Where is the null point located?

## Two charges $+ 1 \cdot {10}^{-} 6$ and $- 4 \cdot {10}^{-} 6$ are separatd by a distance of 2 m. Where is the null point located?

May 6, 2018

$2 m$ from the lesser charge and $4 m$ from the larger charge.

#### Explanation:

We are looking for the point where the force on a test charge, introduced near the 2 given charges, would be zero. At the null point, the attraction of the test charge toward one of the 2 given charges would be equal to the repulsion from the other given charge.

I will choose a one dimensional reference system with the - charge, ${q}_{-}$, at the origin (x=0), and the + charge, ${q}_{+}$, at x = + 2 m.

In the region between the 2 charges, the electric field lines will originate at the + charge and terminate at the - charge. Remember that the electric field lines point in the direction of the force on a positive test charge. Therefore the null point of the electric field must lie outside the charges.

We also know that the null point must lie closer to the lesser charge in order for the magnitudes to cancel- as $F \propto \left(\frac{1}{r} ^ 2\right)$- it decreases as a square over distance. Therefore the coordinate of the null point will have $x > + 2 m$. The point at which the electric field is zero would also be the point (the null point) where the force on a test charge would be zero.

Using Coulomb's law, we can write separate expressions to find the force on a test charge, ${q}_{t}$, due to the two separate charges. Coulomb's Law in formula form:

$F = k \frac{\left({q}_{1}\right) \times \left({q}_{2}\right)}{{r}^{2}}$

Using that to write our separate expressions (see above paragraph) for a null point at x

${F}_{-} = k \frac{\left({q}_{t}\right) \times \left({q}_{-}\right)}{{x}^{2}}$
Note, I am using ${F}_{-}$ to designate the force on the test charge, ${q}_{t}$, due to the negative charge, ${q}_{-}$.
F_+ = k((q_t) times (q_+))/((x-2)^2

The 2 forces on ${q}_{t}$, due individually to ${q}_{-} \mathmr{and} {q}_{+}$, must sum to zero

${F}_{-} + {F}_{+} = 0$.

$k \frac{\left({q}_{t}\right) \times \left({q}_{-}\right)}{{x}^{2}} + k \frac{\left({q}_{t}\right) \times \left({q}_{+}\right)}{{\left(x - 2\right)}^{2}} = 0$

Cancelling where possible:

$\frac{{q}_{-}}{{x}^{2}} + \frac{{q}_{+}}{{\left(x - 2\right)}^{2}} = 0$

Plugging in the charge values:

$\frac{- 4 \times {10}^{-} 6}{{x}^{2}} + \frac{1 \times {10}^{-} 6}{{\left(x - 2\right)}^{2}} = 0$

Some cancelling again, and rearranging,

$\frac{1}{{\left(x - 2\right)}^{2}} = \frac{4}{{x}^{2}}$

This can be turned into a quadratic- but lets make it simple and take the square root of everything, yielding:

$\frac{1}{x - 2} = \frac{2}{x}$

Solving for x:

$x = 2 x - 4$

$x = 4$