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gadje
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Gauss's Law and Superposition of Fields (edited again, something else wrong)
Right. The shape itself has charge Q, so it has charge density \frac{Q}{\frac{4}{3} \pi R^3 - \frac{4}{3} \pi (\frac{R}{2})^3} = \frac{6Q}{7\pi R^3} Let's call this \rho. If it were filled in entirely, then, it would have charge:
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. (V being the volume of the small sphere.)
By Gauss's Law, the field at the surface of this filled sphere would be:
E = \frac{\frac{8}{7}Q}{4 \pi \epsilon_0 R^2} = \frac{2}{7 \pi \epsilon_0 R^2}
Considering now the smaller, negatively charged sphere, this would be carrying a charge of
-\rho V = -\frac{6Q}{7\pi R^3} \cdot \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = -\frac{Q}{7}.
The electric field at its surface would then be \frac{\frac{-Q}{7}}{4 \pi \epsilon (\frac{R}{2})^2} = -\frac{Q}{7 \pi \epsilon _0 R^2}.
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.
Homework Statement
Right. The shape itself has charge Q, so it has charge density \frac{Q}{\frac{4}{3} \pi R^3 - \frac{4}{3} \pi (\frac{R}{2})^3} = \frac{6Q}{7\pi R^3} Let's call this \rho. If it were filled in entirely, then, it would have charge:
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. (V being the volume of the small sphere.)
By Gauss's Law, the field at the surface of this filled sphere would be:
E = \frac{\frac{8}{7}Q}{4 \pi \epsilon_0 R^2} = \frac{2}{7 \pi \epsilon_0 R^2}
Considering now the smaller, negatively charged sphere, this would be carrying a charge of
-\rho V = -\frac{6Q}{7\pi R^3} \cdot \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = -\frac{Q}{7}.
The electric field at its surface would then be \frac{\frac{-Q}{7}}{4 \pi \epsilon (\frac{R}{2})^2} = -\frac{Q}{7 \pi \epsilon _0 R^2}.
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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