# Two particles with positive charges q1 and q2 are separated by a distance s. Along the line...

## Question:

Two particles with positive charges q1 and q2 are separated by a distance s.

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero?

Express your answer in terms of some or all of the variables s, q1, q2 and k.

## The Neutral Point In An Electric Field

According to Coulomb's law the electric field strength at a point at a distance {eq}\displaystyle {r} {/eq} from a point charge {eq}\displaystyle { q} {/eq} is given by,

{eq}\displaystyle { \vec{E}=\frac{1}{4 \pi \epsilon_0}\frac{q}{r^2}\hat{r}} {/eq}.

Here {eq}\displaystyle {\epsilon_0} {/eq} is called the permittivity of free space and {eq}\displaystyle {\frac{1}{4\pi \epsilon_0}=9\times 10^9\ SI\ units} {/eq}.

Like charges repel while unlike charges attract. Each charge will exude an invisible electrical fluid and this fluid supports a force field. It is by recourse to this particular field that charges seemingly act on each other at a distance.

Let the line connecting the two charges be the x-axis. The net field will vanish at that intermediate point between the two positive charges where the electric field strength due to the two individual charges are equal. Let this neutral point be at a distance x from {eq}\displaystyle {q_1} {/eq}. Then it at a distance of s-x from {eq}\displaystyle {q_2} {/eq}. Therefore we have,

{eq}\displaystyle { \frac{q_1}{x^2}=\frac{q_2}{(s-x)^2}} {/eq}

Re-arranging and taking the positive square root,

{eq}\displaystyle { \frac{x}{s-x}=\sqrt{\frac{q_1}{q_2}}} {/eq}

Or,

{eq}\displaystyle {x=\frac{s}{1+\sqrt{\frac{q_2}{q_1}}}} {/eq}.

Note: The negative square root would've yielded an unphysical value greater than s. 