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A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on the xx axis is the net electric field zero? (Be careful to keep track of the direction of the electric field of each particle.)

Answer :

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

[tex]\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}[/tex]

Put the value into the formula

[tex]\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}[/tex]

[tex]\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}[/tex]

[tex]\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}[/tex]

[tex]\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}[/tex]

[tex]\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1[/tex]

[tex]x=0.32\ m[/tex]

Hence, The position on the x axis is 0.32 m.

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