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

1. May 19, 2010

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 $$\frac{Q}{\frac{4}{3} \pi R^3 - \frac{4}{3} \pi (\frac{R}{2})^3} = \frac{6Q}{7\pi R^3}$$ 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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Last edited: May 19, 2010
2. May 19, 2010

### kuruman

You have the correct answer. Note that (2/7) + (-1/7) = 1/7, which is the answer you are meant to get.

Oh, by the way, note that the answer you are meant to get has a 4 in the numerator and in the denominator, so multiply and divide your answer by 4.

3. May 19, 2010