Question #233038
Two beads one carrying charge +q the other +4q are seperated by distance d, that is much greater than the radius of each bead.

Is there any location along the line between them where the electromagnetic field magnitude is zero. If so, what is the distance from the particle carrying +q to the location?

Write answer in term of d(unless there is no answer enter none)
1
Expert's answer
2021-09-04T07:34:57-0400

The electric field produced by the first bead at some distance xx from it is given as follows:


E1=kqx2E_1 = k\dfrac{q}{x^2}

where kk is the constant and xx is the distance from the first bead. The second bead produces the following field at the same point xx:



E2=k4q(dx)2E_2 = k\dfrac{4q}{(d-x)^2}

These fields are directed in opposite direction (in the region between beads), thus, the resulting field is zero if


E1=E2E_1 = E_2

Solving for xx, obtain:


kqx2=k4q(dx)21x2=4(dx)2(dx)2=4x2d22dx+x2=4x23x2+2dxd2=0x=d+4d2+12d26=d+4d6=56dk\dfrac{q}{x^2} = k\dfrac{4q}{(d-x)^2}\\ \dfrac{1}{x^2} = \dfrac{4}{(d-x)^2}\\ (d - x)^2 = 4x^2\\ d^2 - 2dx + x^2 = 4x^2\\ 3x^2 + 2dx - d^2 = 0\\ x = \dfrac{d + \sqrt{4d^2 +12d^2}}{6} = \dfrac{d + 4d}{6} = \dfrac{5}{6}d\\



Answer. 56d\dfrac{5}{6}d.


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