(adsbygoogle = window.adsbygoogle || []).push({}); Gauss's Law and Superposition of Fields (edited again, something else wrong)

1. The problem statement, all variables and given/known data

Right. The shape itself has charge Q, so it has charge density [tex]\frac{Q}{\frac{4}{3} \pi R^3 - \frac{4}{3} \pi (\frac{R}{2})^3} = \frac{6Q}{7\pi R^3}[/tex] Let's call this [tex]\rho[/tex]. If it were filled in entirely, then, it would have charge:

[tex] Q + \rho V = Q + \frac{6Q}{7\pi R^3}\cdot \frac{4}{3}\pi\left(\frac{R}{2}\right)^3 = \frac{8}{7} Q[/tex]. (V being the volume of the small sphere.)

By Gauss's Law, the field at the surface of this filled sphere would be:

[tex] E = \frac{\frac{8}{7}Q}{4 \pi \epsilon_0 R^2} = \frac{2}{7 \pi \epsilon_0 R^2} [/tex]

Considering now the smaller, negatively charged sphere, this would be carrying a charge of

[tex]-\rho V = -\frac{6Q}{7\pi R^3} \cdot \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = -\frac{Q}{7}[/tex].

The electric field at its surface would then be [tex]\frac{\frac{-Q}{7}}{4 \pi \epsilon (\frac{R}{2})^2} = -\frac{Q}{7 \pi \epsilon _0 R^2}[/tex].

The electric field at P of the shape, then, is the sum of the individual fields of the filled sphere and the negatively charged sphere - I don't get the answer I'm meant to, and I can't see what I've done wrong. It's probably something really stupid.

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# Homework Help: Gauss's Law and Superposition of Fields

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