- By my understanding, the electric field in the surface integral expression for Gauss's Law represents the total electric field and any point on a closed Gaussian surface. However, when we employ Gauss's Law to find the electric field within a uniform charge distribution, we ignore any charges outside the Gaussian surface and look only at the charge enclosed. My question is - if the electric field expression in Gauss's Law represents the total E field at a point, why do we only consider enclosed charges and not those outside a defined surface?
For example, consider a charge distribution of uniform charge density $ρ$, total charge $Q$, and radius $R$. When asked to find $\vec E$ at radius $ r<R$, we can solve by the following. $$\int \vec E \cdot dA = \frac{Q_{enclosed}}{ε_0} $$ where $$ Q_{enclosed} = ρV = \frac{r^3}{R^3}Q $$ $$E = \frac{k_eQ}{R^3}r $$ This solution would have been the same regardless of whether charges external to the Gaussian surface of radius $r$ had existed. In other words, external charges seemingly do not affect the magnitude of $ \vec E$ in this case. How can this be possible?
- I am trying to determine the direction of electric field within a uniformly charged, insulated sphere of radius $R$ and total positive charge $Q$
My solution: knowing that $V = -\int \vec E \cdot dR $, I find the expression for potential as a function of radius $ r$ and define electric field direction as that opposite to increasing potential.
$$V(r) = V(R) - \int_R^r \vec E \cdot dR$$ where, as previously determined, $ E$ for $ r<R$ is $\frac{k_eQ}{r^3}r$ .Thus: $$ V(r) = \frac{k_eQ}{R} - \frac{k_eQ}{2R^3}(r^2-R^2)$$
As $r$ increases, $V(r)$decreases and therefore $\vec E$ is directed radially outward. My question - is there an easier method to visualize why?
