Consider a system of three point charges ont the x axis. Charge 1 is at x=0, charge 2 is at x=0.20m, and charge 3 is at x = 0.40m. In addition, the charges have the following values: q1 = -19uC, q2=q3=+19uC. Find the the point where E =0 between x =0.20m?

Find the point where E =0 between x = 0.20m and x = 0.40m

Mar 22, 2017

The situation as described in the problem is shown in the figure.
Let the position where the system charges will have net net intensity $E = 0$ is the point $N$ which is $x$ m from ${q}_{2}$.

Intensity at N due to ${q}_{1}$along $\vec{N O} = {E}_{1} = {k}_{c} \times \frac{19 \times {10}^{-} 6}{0.2 + x} ^ 2 N {C}^{-} 1$.

Intensity at N due to ${q}_{3}$along $\vec{N O} = {E}_{3} = {k}_{c} \times \frac{19 \times {10}^{-} 6}{0.2 - x} ^ 2 N {C}^{-} 1$.

Intensity at N due to ${q}_{2}$along $\vec{N Q} = {E}_{2} = {k}_{c} \times \frac{19 \times {10}^{-} 6}{x} ^ 2 N {C}^{-} 1$.

where Coulomb's constant, ${k}_{c} = 9 \times {10}^{9} N {m}^{2} \text{/} {C}^{2}$.

So considering the equilibrium we can write

${E}_{1} + {E}_{3} = {E}_{2}$

$\implies {k}_{c} \times \frac{19 \times {10}^{-} 6}{0.2 + x} ^ 2 + {k}_{c} \times \frac{19 \times {10}^{-} 6}{0.2 - x} ^ 2 = {k}_{c} \times \frac{19 \times {10}^{-} 6}{x} ^ 2$

So we have

$\frac{1}{0.2 + x} ^ 2 + \frac{1}{0.2 - x} ^ 2 = \frac{1}{x} ^ 2$

$\implies \frac{2 \times {0.2}^{2} + 2 \times {x}^{2}}{0.04 - {x}^{2}} ^ 2 = \frac{1}{x} ^ 2$

$\implies \frac{0.08 + 2 {x}^{2}}{0.04 - {x}^{2}} ^ 2 = \frac{1}{x} ^ 2$

$\implies \left(0.08 + 2 {x}^{2}\right) {x}^{2} = {\left(0.04 - {x}^{2}\right)}^{2}$

$\implies 2 {x}^{4} + 0.08 {x}^{2} = {x}^{4} - 0.08 {x}^{2} + 16 \times {10}^{-} 4$

$\implies {x}^{4} + 2 \times 0.08 {x}^{2} + {\left(0.08\right)}^{2} = 16 \times {10}^{-} 4 + {\left(0.08\right)}^{2}$

$\implies {\left({x}^{2} + 0.08\right)}^{2} = 80 \times {10}^{-} 4$

$\implies {x}^{2} = \sqrt{80 \times {10}^{-} 4} - 0.08 \approx 0.00944$

$\implies x \approx \sqrt{0.00944} \approx 0.09 m$

So location of $N$ on x-axis where $E = 0$ is color(red)((0.29,0)