Question

Two charges are placed on the x axis, +5.00 μC at the origin and −10.0 μC at x = 10. 0 cm . (a) Find the electric field on the x axis at x = 6. 00 cm . (b) At what point(s) on the x axis is the electric field zero?

Answer 1

Answer:

a) $E=1.875\times 10^{-6}\ N.C^{-1}$

b) $x_0=-0.06\ m$ towards the negative side of the origin.

Explanation:

Given:

charges and their respective positions:

$Q_1=5\times 10^{-6}\ C$           at           $x_1=0$

$Q_2=-10\times 10^{-6}\ C$      at           $x_2=0.1\ m$

a)

Now the electric field at point x=0.06\ m will be:

(As we know that the electric field is away from the positive charge and towards the negative charge.)

$E=\frac{1}{4\pi\epsilon_0} \times (\frac{Q_1}{x_1-x}+\frac{Q_2} {x_2-x} )$

$E=9\times 10^{-9} \times (\frac{5\times 10^{-6}}{0.06}+\frac{5\times 10^{-6}} {0.1-0.06} )$

$E=1.875\times 10^{-6}\ N.C^{-1}$

b)

Now since the charges are opposite in nature the neutral point will not lie between the charges and hence the neutral point lies close to the smaller charge and beyond the two charges.

$E_1=E_2$

$\frac{1}{4\pi\epsilon_0} \times \frac{Q_1}{r} =\frac{1}{4\pi\epsilon_0} \times \frac{Q_2}{r+0.06}$

$\frac{5\times 10^{-6}}{r} =\frac{10\times 10^{-6}}{r+0.06}$

$r=0.06\ m$ towards the negative side of the origin.