Simple isn't it? #3

Coulomb's Law states that the force is $\frac{k q_1 q_2}{r^2}$ . Since the charges and Coulomb's constant are already known, we need to find the expected value of $\frac{1}{r^2}$ , for two randomly chosen points, one on the first bar, and one on the second.

Suppose the bars are on a number line, with the first bar beginning at $0$ and ending at $16$ . The second bar then starts at $20$ and ends at $36$ .

Let the first point be located at $x$ . Then, the expected value is $\frac{1}{16}\int_{20}^{36} \frac{1}{(y-x)^2}dy=\frac{1}{16}(-\frac{1}{36-x}-(-\frac{1}{20-x}))=\frac{1}{16}(\frac{1}{20-x}-\frac{1}{36-x})$ . Now the expected value of this, as $x$ ranges from $0$ to $16$ , is $\frac{1}{256}\int_0^{16}{\frac{1}{20-x}-\frac{1}{36-x}}dx=\frac{1}{256}((-\ln{\frac{20-16}{36-16}})-(-\ln{\frac{20-0}{36-0}}))=\frac{1}{256}\ln(\frac{20}{36}\times\frac{20}{4})=\frac{1}{256}\ln{\frac{400}{144}}=0.00399$ .

Finally, by Coulomb's Law, the answer is $40\times40\times(9\times10^9)\times{0.00399}=\boxed{5.75\times10^{10}}$

Nice question but not worth level 5. solution

By coulomb's law, F = $k\frac{q_1q_2}{r^2}$ And the distance between the inner edges of the two rods is 4m.Considering uniform charge distribution, charge per unit length is $\frac{40}{16}$ . Considering a very small element dx on any one of the rods at a distance x from the inner edge, the total force exerted on it by every small element dy at a distance y form the inner edge on the other rod is, $k\frac{40^2}{16^2}\frac{dx dy}{(4+x+y)^2}$ Integrating this throughout the length of the first and then the second rod, $k\frac{40^2}{16^2}\int_{0}^{16}dy\int_{0}^{16}\frac{dx}{(4+x+y)^2}$ . Solving the integration and substituting the value of k , the answer is $\boxed{5.7468E10}$ .

Here is the straight forward bulldozer way of solving the problem. We start by finding the electric field contribution $\overrightarrow{E}(R)$ of a thin rod of length $L$ at a point $R$ . We assume that the rod has a uniform charge distribution $\lambda$ .

To do so we consider a infinitesimal charge $dq$ on the rod which is a distance $x$ away and its contribution to the red point a distance $R$ away from the origin. If the rod is uniformly charged then $dq = \lambda dx$ . ( $dx$ is an infinitesimal distance away from $x$ ). Thus from coloumb's law

$\overrightarrow{dE}(R) = \frac{k\lambda dx}{(R-x)^2}$ We then integrate from $0$ to $L$ , to obtain the electric field due to a rod at a point $R$ away from its furthest edge: $\overrightarrow{E}(R) = \int_{0}^{L} \overrightarrow{dE}(R) = k\lambda(\frac{1}{R-L} - \frac{1}{R})$

Now consider two rods one with its edge at the $(0,0)$ and the other $d$ away from the origin. The infinitesimal force contribution due to the left most rod on the right one is $d\overrightarrow{F}(x) = dq\overrightarrow{E}(x) = \lambda_2 dx \overrightarrow{E}(x) = \frac{kQ1Q2}{l_1 l_2} \int_{d}^{d+l_2} \frac{1}{x-l_1} - \frac{1}{x} dx$

Finally, a little python script to do the integral.(Not that its hard, It just because Im lazy and numerical integration is cool)

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from scipy import integrate
l1 = l2 = 16
Q1 = Q2 = 40
d , k = 20 , 9e9
def f(x):
    return 1.0/(x-l1) - 1.0/(x)
print (k*Q1*Q2)/(l1 * l2) * integrate.quad( f , d , d + l2 )[0]

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>>> 
57467882673.7

Thoughts

I am curious if such a configuration is possible. $40C$ seems like an awful lot of charge. I know air breaks down and becomes a conductor when the electric field is too big. Even assuming the rods are in a vaccum , I wonder if the charge put in the metal exceeds the maximum possible surface charge density and the rods will start exhibiting field electron emission . Does anyone know how to calculate this? I might be wrong.