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Caelan

Answered question

2020-11-22

In figure, determine the point at which the total electric field is zero. Find ____m to the left of $- 2.5 \times 10^{- 6}$ C charge.

Answer & Explanation

Tuthornt

Skilled2020-11-23Added 107 answers

Take origin at -2.5 micro Coulomb charge and both the chargeson X-axis for convenience
Let the point where the field is zero to the left of -2.5muCcharge be at x meters from it

$E_{1} = \frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}} = \frac{1}{4 π ϵ_{0}} \frac{2.5 \times 10^{- 6}}{x^{2}}$
to wards right along x-axis at the point P
$E_{2} = \frac{1}{4 π ϵ_{0}} \frac{6 \times 10^{- 6}}{(1 + x)^{2}}$

towards left along x-axis at tthe point P
$E_{1} = E_{2}$
So that net field is zero
$\frac{1}{4 π ϵ_{0}} \frac{2.5 \times 10^{- 6}}{x^{2}} = \frac{1}{4 π ϵ_{0}} \frac{6 \times 10^{- 6}}{{(1 + x)}^{2}}$
$\frac{2.5}{x^{2}} = \frac{6}{{(1 + x)}^{2}}$
$7 x^{2} - 10 x - 5 = 0$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{10 \pm \sqrt{100 + 140}}{14}$
$x = \frac{25.49}{14} or \frac{- 5.49}{14}$
$x = + 1.82 m$ or $x = - 0.39 m$
Answer is 0.39 m to the left of -2.5 microcoulomb charge

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