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We know that work done $W$ to assemble $q_1$,...,$q_n$ point charges is $$ W = \frac{1}{2} \sum_{i = 1}^{n} q_i V(r_i) \tag{I} $$

Now for the continous charge distribution with charge density $k$, we find $W$ by integrating over appropriate region, $$ W = \frac{1}{2}\int \rho V \, d\tau \tag{II} $$

Now using Gauss law , $\rho = \epsilon_{0} \nabla\cdot\vec{E})$ and vector calculus identities we find that $$ W = \frac{\epsilon_0}{2} \int_{\text{all space}} E^2 \, d\tau \tag{III} $$

Now my doubt is this expression is always positive while the former expression (in case of discrete case) could take negative values as well. How this is possible even though continous case is considered as limit of finite cases?

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In equation (I) only the work spent to bring whole point charges (with finite charges) from infinity to the neighbourhood of each other is computed, but not the work spent to assemble each individual point charge, which is (theoretically) positive infinite for each point charge, because of the infinite repulsion overcome while compressing some finite amount of charge into a single geometric point. So the equation (II) and (III) count for that work and therefore they are positive.

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A part of the problem is not differentiating between the work and the energy (which is defined via work done by removing charge to infinity or bringing it from infinity - depending on the one's taste).

Further, the potential in the first expression may originate either from the charges of interest or from the external charges, so the energy could be either negative or positive. However the second and the third equations seem to suggest a more specific problem of calculating the energy of the charge distribution due to the charges of this distribution itself. This energy is positive, since the charges repel each other and the energy of the system could only decrease, if they move further apart.

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