In the figure particle 1 of charge q 1 = -7.73 q and particle 2 of charge q 2 =

Question

In the figure particle 1 of charge q1 = -7.73q and particle 2 of charge q2 = +3.13q are fixed to an x axis. As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero?

I got this equation, but not sure how to solve for x: (x-L)/(x) =?(3.13/7.73)

Or is this equation even correct?

In the figure particle 1 of charge q1 = -7.73q and particle 2 of charge q2 = +3.13q are fixed to an x axis. As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero? I got this equation, but not sure how to solve for x: (x-L)/(x) =?(3.13/7.73) Or is this equation even correct?

Explanation / Answer

let Q1 = +q

and Q2 = -2*q

here |Q1| > |Q2|

so, the point where Enet = 0 shold be close to Q2 and it should be right side.

let at a dustance x from the origin Enet = 0

E1 = E2

k*Q1/(L+x)^2 = k*Q2/(x)^2

7.73/(L+x)^2 = 3.13/(x)^2

x/(L+x) = sqrt(3.13/7.73)

x = (L+x)*0.636

x = 0.636*L + x*0.636

x = 0.636*L/(1-0.636)

x = 1.75*L

so, at x = L + 1.75*L = 2.75*L Enet = 0

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