Consider a charge of +q at (1, 0) m, and -q at (-1, 0)m. Recall that the electri

Question

Consider a charge of +q at (1, 0) m, and -q at (-1, 0)m. Recall that the electric field of a point charge is: E + k q_0/r^2 r Calculate the field strength at a point (0, y) m as a function of y. Look for symmetries of the two charges to cancel some components of the field, and simplify the vectors. Consider how the field strength on the y-axis varies with y as y gets much larger than the distance between the charges You might find that the field drops off as 1/s^2, and if you do so, you might be correct.

Explanation / Answer

field due to +q at (0,y):

field direction is away from the charge.

vector along the field=(0,y)-(1,0)=(-1,y)

distance=sqrt(1+y^2)

unit vector=(-1,y)/sqrt(1+y^2)

field magnitude=k*q/distance^2=k*q/(1+y^2)

in vector term, field=E1=(k*q/(1+y^2))*(-1,y)/sqrt(1+y^2)

field due to -q at (0,y):

field direction is towards the charge.

vector along the field=(-1,0)-(0,y)=(-1,-y)

distance=sqrt(1+y^2)

unit vector=(-1,-y)/sqrt(1+y^2)

field magnitude=k*q/distance^2=k*q/(1+y^2)

in vector term, field=E2=(k*q/(1+y^2))*(-1,-y)/sqrt(1+y^2)

net field=E1+E2=(2*k*q/(1+y^2)^1.5)*(-1,0)

so field strength is 2*k*q/(1+y^2)^1.5

and is along -ve x axis

as y becomes larger, 1+y^2=y^2 approximately

then field strength=2*k*q/(y^2)^1.5=2*k*q/y^3

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