# Simple electrostatics problem

1. Feb 21, 2008

### jacobrhcp

[SOLVED] simple electrostatics problem

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

a cylinder, the inhomogenous E field is given by $$E = E_{0} (\frac{r}{R})^{3}$$

the assignment was to calculate Q of a piece of the cylinder of length L in two different ways, by gauss law in integral form and in differential form

3. The attempt at a solution

$$\oint E dA = \frac{Q}{e_{0}}$$

so $$Q = 2\pi e_{0} L E_{0} R$$

the other way around, $$\nabla E = \frac{\rho}{e_{0}}$$

so $$\rho = e_{0} E_{0} 3 \frac{r^{2}}{R^{3}}$$

so $$Q = e_{0} E_{0} 3 \oint \frac{r^{2}}{R^{3}} r dr d\theta dl = 2 \pi e_{0} E_{0} \frac{3}{4} L R$$

which is a factor 3/4 different from my other expression for Q

where did I go wrong?

Last edited: Feb 21, 2008
2. Feb 21, 2008

### Dick

Review the form of the divergence in cylindrical coordinates. The non-vanishing term is (d/dr+1/r)E_r. That changes your '3' in rho to a '4'.