MHB Electrostatic self-potential energy

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The electrostatic self-potential energy of a spherical charge distribution with uniform charge density ρ and radius R is derived from the work needed to increase the sphere's radius from r to r+dr. The charge element dq is expressed as 4πρr², while the potential V is calculated as (4πρk r²)/3, with k being Coulomb's Law constant. By multiplying the expressions for work dW and integrating from 0 to R, the self-potential energy is found to be (16/15)π²ρ²kR⁵. This formula quantifies the energy associated with expanding the radius of the charge distribution. Understanding this concept is crucial for applications in electrostatics and related fields.
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Find the electrostatic self potential energy of a spherical charge distribution with charge density $$\rho$$ and radius $$R$$. The self potential energy is the work required to increase the radius of the sphere from $$r$$ to $$r+dr$$.
 
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Wow, I forgot this was even here.
I'll post a solution, since I don't like unanswered threads.
The electrostatic self-potential energy is the amount of work required to increase the radius of a spherically symmetric charge distribution from r to r+dr. From the definition of voltage, $$dW=Vdq$$. The charge dq is $$4 \pi \rho r^2$$. The potential V is $$\frac{kq}{r}=\frac{4 \pi \rho k r^3}{3r}=\frac{4 \pi \rho k r^2}{3}$$. $$k$$ is the Coulomb's Law constant, $$k=\frac{1}{4 \pi \epsilon_0}$$. Multiplying these two expressions, we have $$\frac{16 \pi^2 \rho^2 k r^4}{3}$$. Integrating from 0 to R (increasing the radius from 0 to R by putting on more and more shells), we have
$$\frac{16}{3} \pi^2 \rho^2 k \int_0^R r^4 dr=\frac{16}{15} \pi^2 \rho^2 k R^5[/math] as our answer.
 
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