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Here's the crux of the problem: break it up. As an example, let's find the forces on charge 1. First, we have to find the force on charge 1 ($q_1$) by charge 2 ($q_2$). We then use Coulomb's law, $$F=k\frac{q_1q_2}{r^2}$$ We now have to find $r$. Using the Pythagorean theorem, $r^2=r_x^2+r_y^2$, so $r=\sqrt{r_x^2+r_y^2}$. If we know $x_1$, $x_2$, $y_1$, and ...


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$\oint E\cdot dS = \frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV $ if $\frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV$ then $\frac{\rho}{\epsilon} = \nabla \cdot E$ if $\rho=0$ then $\frac{\rho}{\epsilon} = 0 = \nabla \cdot E$ Is this what your looking for? $\rho$ would be zero say, ...


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The answer to the main question is YES. Two electrons will "touch" each other when their centers are at a separation equal to one electron diameter. Since the diameter of an electron is not zero, an infinite amount of energy, is not required to make them "touch." With a (calculated) electron diameter = to $2.82 \times 10^{-15}$ m, the required energy can be ...



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