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not in book

.Two positive charges are fixed in place. A 5
.0mC charge is
placed 2m to theleft of the origin and a 3 .0mC charge is
placed1m to the left of the origin. At
what point(s) is theelectric field zero?

Respuesta :

Answer:

[tex]x=2.4365\ m[/tex]

and

[tex]x=-1.4365\ m[/tex]

Explanation:

Given:

  • first charge, [tex]q_1=5\times 10^{-3}\ C[/tex]
  • second charge, [tex]q_2=3\times 10^{-3}\ C[/tex]
  • position of first charge, [tex]x_1=-2\ m[/tex]
  • position of second charge, [tex]x_2=-1\ m[/tex]

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

[tex]E_1=E_2[/tex]

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

[tex]\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}[/tex]

[tex]\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}[/tex]

[tex]3(r^2+1+2r)=5r^2[/tex]

[tex]2r^2-6r-3=0[/tex]

[tex]r=3.4365 \&\ r=-0.4365[/tex]

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

[tex]x=-1+3.4365=2.4365\ m[/tex]

and

[tex]x=-1-0.4365=-1.4365\ m[/tex]