Thunderstorm Discharge modeled as Electric Dipole

[Nyfinscyf]

Homework Statement

An electric dipole with -q at the clouds of height h, and +q beneath the surface with height -h.
Given q=200 C, and h=6000m
Electric discharge occurs with electric field of 3000 V/m near ground level, and 200 V/m above the clouds.

Find the ranges where discharges are likely.

Homework Equations

\vec{E}=\frac{kq}{r^{2}}\hat{r}
k=\frac{1}{4\pi \epsilon_{0}}
E=\left| \vec{E} \right|=\frac{k \left| q \right|}{r^{2}}

The Attempt at a Solution

I found an expression for the electric field at some point a distance z away from the ground
E=E_1 + E_2 = 2kq\frac{z^2 + h^2}{(z^2 - h^2)^2}
plugging in z=0 (to find electric field at ground level) you get:
E=2kq\frac{1}{h^2}≈100 000 V/m
And the closer you move to the clouds the larger the electric field gets, so the first range of discharges is simply 0 < z < h.
Now the second ranges of likely discharges above the clouds can come from the same expression but I can't simplify it to find an actual distance.
I tried setting E=200 but I can't simplify it. plugging in large numbers approx 130 000 m as z gives an electric field less than 200 V/m.

I don't think I'm approaching this problem correctly. Can anyone help me out?

 

[Greg Bernhardt]

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?

 

[gneill]

Homework Statement

An electric dipole with -q at the clouds of height h, and +q beneath the surface with height -h.
Given q=200 C, and h=6000m
Electric discharge occurs with electric field of 3000 V/m near ground level, and 200 V/m above the clouds.

Find the ranges where discharges are likely.

Homework Equations

\vec{E}=\frac{kq}{r^{2}}\hat{r}
k=\frac{1}{4\pi \epsilon_{0}}
E=\left| \vec{E} \right|=\frac{k \left| q \right|}{r^{2}}

The Attempt at a Solution

I found an expression for the electric field at some point a distance z away from the ground
E=E_1 + E_2 = 2kq\frac{z^2 + h^2}{(z^2 - h^2)^2}

Hi Nyfinscyf. I think you should check the denominator in your expression above. When z lies between the ground and cloud layer, as it gets further from the lower charge it will get closer to the upper charge. Surely one distance is (h + z) and the other (h - z), which multiplied together won't give you your denominator.

plugging in z=0 (to find electric field at ground level) you get:
E=2kq\frac{1}{h^2}≈100 000 V/m
And the closer you move to the clouds the larger the electric field gets, so the first range of discharges is simply 0 < z < h.

Looks good. Note the denominator issue above disappears when z = 0!

Now the second ranges of likely discharges above the clouds can come from the same expression but I can't simplify it to find an actual distance.
I tried setting E=200 but I can't simplify it. plugging in large numbers approx 130 000 m as z gives an electric field less than 200 V/m.

I don't think I'm approaching this problem correctly. Can anyone help me out?

Write a new expression for the region above the clouds. In that case the distance to both charges increases as this new z increases. Be sure to take note of the field directions for the individual charges in that region.

 

[Nyfinscyf]

I wrote another expression for above the clouds, where z is the distance from the clouds. The E field from the charges are in opposite directions which is why there is a negative in the expression below:
E_{above}=E_{+}+E_{-}=kq(\frac{1}{(2h+z)^2} - \frac{1}{z^2})
which simplifies to
E_{above}=-4kqh{\frac{h+z}{(2hz+z^2)^2}}
Which might not be a more simple expression

Now with this approximately 55km above the clouds is where the E field is less than 200 V/m, this was found just by plugging in numbers. I figured there is a more accurate way to do it.

The field is negative because it's getting the most influence from the negative charge at the clouds.

Surely one distance is (h + z) and the other (h - z), which multiplied together won't give you your denominator.

\frac{1}{(h+z)^2} + \frac{1}{(h-z)^2}=\frac{(h-z)^2+(h+z)^2}{(h+z)^2(h-z)^2}=\frac{h^2+z^2-2hz+h^2+z^2+2hz}{((h+z)(h-z))^2}=2\frac{h^2+z^2}{(h^2-z^2)^2}

Same as what I wrote, except denom is reversed which doesn't matter since it's squared anyways.

 

[gneill]

I wrote another expression for above the clouds, where z is the distance from the clouds. The E field from the charges are in opposite directions which is why there is a negative in the expression below:
E_{above}=E_{+}+E_{-}=kq(\frac{1}{(2h+z)^2} - \frac{1}{z^2})
which simplifies to
E_{above}=-4kqh{\frac{h+z}{(2hz+z^2)^2}}
Which might not be a more simple expression

Not necessarily more simple, but it is correct for the region it applies to. The other expression does you had did not apply to the region above the clouds due to the sign reversal of the cloud field as you "pass through" it on the way upwards.

Now with this approximately 55km above the clouds is where the E field is less than 200 V/m, this was found just by plugging in numbers. I figured there is a more accurate way to do it.

Looks about right. Since you end up with a nasty fourth order polynomial a numerical approach is probably the best way to go.

The field is negative because it's getting the most influence from the negative charge at the clouds.


\frac{1}{(h+z)^2} + \frac{1}{(h-z)^2}=\frac{(h-z)^2+(h+z)^2}{(h+z)^2(h-z)^2}=\frac{h^2+z^2-2hz+h^2+z^2+2hz}{((h+z)(h-z))^2}=2\frac{h^2+z^2}{(h^2-z^2)^2}

Same as what I wrote, except denom is reversed which doesn't matter since it's squared anyways.

Yeah, okay. I must have been seeing things

This equation can be reduced to a quadratic in $z^2$. That is, make a substitution say $x = z^2$, then solve for x.

 

[Nyfinscyf]

Thanks for your help gneill.
I'm happy with the answer now.