Please show all your works and answer correctly. Thanks. Given two charges, 3Q a

Question

Please show all your works and answer correctly. Thanks.

Given two charges, 3Q at x = 0.0 mm and 5Q at x = 2.0 mm:

a) Assuming V = 0 at , are there any other points along the x-axis where V = 0? If so, find it/them.

b) What if the 3Q charge is instead -3Q? Again, find any and all points along the x-axis where V = 0.

c) In HW 5 you found that E = 0 occurred at x = +0.87 mm when both charges were positive and at x = -6.87 mm when the 3Q charge was negative. Why does E = 0 not correspond to V = 0? Which do you think is more physically meaningful? Why?

Explanation / Answer

part a:

assumption: V=0 at infinity.

potential at a point which is at a distance d from a charge Q is given by k*Q/d

so as 3*Q and 5*Q are both positive, only at x=inifnity or x=-inifnity, V will be zero.

part b:

if 3*Q charge is replaced with -3*Q, as potential is directly proportional to charge /distance, to balance potential due to one another, the point should be closer to -3*Q as compared to 5*Q.

so it can either happen to the left of -3*Q (when x<0) or in between -3*Q and 5*Q. (for 0<x<2)

for x<0:

let at x=-d, potential is zero.

total potential =0

=>(-3*Q/d)+(5*Q/(d+2))=0

==>5/(d+2)=3/d

==>5*d=3*d+6

==>2*d=6

==>d=3 mm

so at x=-3 mm,V=0

for 0<x<2:

total potential =0

==>(-3*Q/x)+(5*Q/(2-x))=0

==>3/x=5/(2-x)

==>6-3*x=5*x

==>8*x=6

==>x=0.75

hence at x=0.75 mm,V=0

so there are two points where V=0.

those are:

x=-3 mm

x=0.75 mm

part c:

V is a relative function where as E is absolute function.

i.e. V does not have any meaning if there is no referrence point where V=0

so by chosing our referrence point accordingly, we can have V=0 at the point where E=0.

physically electric field is more meaning ful as it is an absolute measurement.

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