Continuous Charge Distributions

discrete point charges

• (like the infinite long charge where we take the integral)

  • be careful with vector nature

• the summation is $\vec{E}={\sum }_{i}k\frac{Q_i}{r_1^2}{\hat{r}}_i$

  • But we need the integral $\vec{E}=\int k\frac{dq}{r^2}\hat{r}$

    • WHAT DOES THIS MEAN

      • conceptually is not different!
      • integrating over all charges ($dq$)
      • $\vec{r}$ is the vector from $dq$ to the point at which $\vec{E}$ is defined

    • Example

      • we have a wire that is charged with Q uniformly, and has a linear density of $\lambda =\frac{Q}{L}$ and we find the point P at a arbitrary point
      • 

• Example 1

  • a straight wire of length 2L carries a uniform linear charge density ${\rho }_e$ ;determine the electric field $\vec{E}$at point P, a distance r from the wire along the perpendicular bisector
  • 
    • $R^{\prime }$is length of the distance from L
      • $dz$ is the varible of integration
    • by symmetry, $\vec{E}=\hat{r}{E}_r$
      • $R^{\prime }=\sqrt{R^2+z^{\prime 2}}$
      • $\cos \alpha =\frac{r}{R^{\prime }}$
    • Solving for $E_r$
      • $E_r=\frac{1}{4\pi {ϵ}_0}\int \frac{dq\ast \cos \alpha }{R^{\prime 2}}dx$

• Example 2

  • The electri field at point P due to a thin uniformly charged rod located on the z axis givien by $\vec{E}=e_z\hat{k}+E_R\hat{R}$

$$\begin{array}{r}{E}_z=\frac{k\lambda }{R}(\sin {\theta }_2-\sin {\theta }_1)\end{array}$$