Question

Two particles with positive charges q1 and q2 are separated by a distance s. Part A 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 =14πϵ0. If your answer is difficult to enter, consider simplifying it, as it can be made relatively simple with some work.

Answer 1

Answer:

$x=s*\frac{q_{1} +-\sqrt{q_{1}q_{2}}}{q_{1}-q_{2}}$

we choose the sign, in order to x<s

Explanation:

x=distance from q1

$E_{1} =E_{2}$

$k*q_{1}/x^{2}=k*q_{2}/(s-x)^2}$

$q_{1}*(s^{2}-2sx+x^{2})-q_{2}x^{2}=0$

$(q_{1}-q_{2})x^{2}-2sq_{1}x+q_{1}*s^{2}=0$

this is a quadratic equation:

$x=\frac{-b+-\sqrt{b^{2}-4ac} }{2a}=\frac{2sq_{1} +-\sqrt{(-2sq_{1})^{2}-4(q_{1}-q_{2})q_{1}s^{2}}} {2(q_{1}-q_{2})}$

$x=s*\frac{q_{1} +-\sqrt{q_{1}q_{2}}}{q_{1}-q_{2}}$