electric field equals zero

TX123 · 15-01-2007 8:12pm #1

if you have two +ve charges Q1 and Q2 seperatdd by s then when is then at what distance from the charge Q1 is the total electric field from the two charges zero.
. How can this be expressed. If your looking for the variable x then can you say that this distance is Sqrt 2s or something. does anyone know?

dudara · 16-01-2007 8:29pm

If the charges are of equal magnitude, then the field is zero halfway between them.

A +ve charge emits a radial field, so if you are situated between the two charges, then you'll experience two fields coming at you from opposite directions. The point at which the net field is zero is where the two fields are equal and opposite.

If the distance between them is d, and the zero-point is at a distance s from Q1, then it's d-s from Q2.

Put these distances into the expressions for electric fields for each charge and then place them equal to each other and solve for s.

TX123 · 16-01-2007 9:59pm

does s then equal to 2d or sqrt2d

dudara · 16-01-2007 10:49pm

The standard formula is Q/(4*PI*epsilon*r^2), where r is the distance from the charge.

If d is the separation between the charges, then the zero-point will be s away from one point and (d-s) away from the other. Set r = s for one and r=(d-s) for the other and put them equal.

Q1/(4*PI*epsilon*s^2) = Q2/(4*PI*epsilon*(d-s)^2)

Now solve for s.

TX123 · 22-01-2007 9:55pm

you end up with something like this.

sqrt Q1/(SqrtQ1 +SqrtQ2) *s

does that sound right ?

dudara · 22-01-2007 11:03pm

If we assume that we're at a distance s from Q1 and a distance (d-s) from Q2, then we get

s = (d*a)/(1+a) where a = SQRT(Q1/Q2)

TX123 · 23-01-2007 8:31pm

i think your wrong there.

dudara · 24-01-2007 12:18am

I don't have access to software at the moment in which I can write the derivation, but if you throw up yours here, I'll look at it.

planck2 · 24-01-2007 11:23pm

If you have a positive charge Q1 and another positive charge Q2 both of which have a non zero E field. how do the E-fields of the charges seperated by a distance s' vanish at every point in space. The new field En = E1 + E2, but E2 is simply cE1, c= Q2/Q1. So the new field En can never vanish.

Also the only components of the fields that vanish are those along the line joining the charges, but this only happens if Q1=Q2. And remember that like charges repel.

dudara · 26-01-2007 12:16am

planck2 wrote:

If you have a positive charge Q1 and another positive charge Q2 both of which have a non zero E field. how do the E-fields of the charges seperated by a distance s' vanish at every point in space. The new field En = E1 + E2, but E2 is simply cE1, c= Q2/Q1. So the new field En can never vanish.

Don't forget that it's not just a matter of scaling the charges, but the distances as well.

Also the only components of the fields that vanish are those along the line joining the charges

I think that there are an infinite number of places where the electric field is zero, but the easiest one to work out is on the line joing the charges. Otherwise, it's more complicated vectors.

but this only happens if Q1=Q2

Not necessarily, when they're unequal is just means that the zeropoint won't be equidistant from both.

And of course, the electric field is always zero at infinity.

planck2 · 26-01-2007 1:46am

but think about it, if Q1 =Q2 then for En to vanish everywhere we must have E1 +E2 = 0 everywhere. then we have E1 = - E2, but this is a contradiction because the charges Q1 and Q2 are identical. the only point where the field En vanishes is the midpoint of the line joing the centre of the charges and nowhere else. And if Q1 = cQ2, c not equal to 1, the distance along the line joing them is only modified by a constant.

And yes the field is zero at infinity

electric field equals zero

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