1. Break the charge distribution into tiny pieces ($dq$). 2. Calculate the electric field contribution ($dE$) from each $dq$. 3. Express $dq$ and $r$ in terms of appropriate coordinates. 4. Integrate all the contributions to find the total electric field ($\vec{E} = \int d\vec{E}$).

1. Calculate the electric field due to each individual charge distribution separately. 2. Express each electric field as a vector, including both magnitude and direction. 3. Add the electric field vectors together to find the net electric field.

1. Identify the type of symmetry (spherical, cylindrical, planar). 2. Use symmetry to determine the direction of the electric field. 3. Use symmetry to identify points where the electric field is zero. 4. Simplify the integral by eliminating components of the electric field that cancel due to symmetry.

1. Define the linear charge density $\lambda = Q/L$. 2. Express $dq = \lambda dx$. 3. Determine the distance $r$ from $dq$ to the point where the field is being calculated. 4. Set up and evaluate the integral $\vec{E}=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r^{2}} \hat{r}$.

1. Define the linear charge density $\lambda = Q/(2\pi R)$. 2. Express $dq = \lambda dl = \lambda R d\theta$. 3. Determine the distance $r = \sqrt{x^2 + R^2}$ from $dq$ to the point on the axis. 4. Recognize that only the x-component of the electric field survives due to symmetry. 5. Set up and evaluate the integral $E_x = \int dE_x = \int dE \cos\theta = \frac{1}{4 \pi \varepsilon_{0}} \int \frac{dq}{r^{2}} \frac{x}{r}$.

Spherical: Field radiates from a central point. Cylindrical: Field radiates from a central axis. Planar: Field is uniform and perpendicular to the plane. All simplify field direction determination. Spherical and Cylindrical fields decrease with distance; Planar field is constant.

Integration: Directly sums contributions from infinitesimal charges, applicable to any distribution. Gauss's Law: Relates electric flux to enclosed charge, simpler for highly symmetric situations (spherical, cylindrical, planar).

The electric field decreases proportionally to the inverse of the distance ($1/r$).

Increasing the charge density increases the magnitude of the electric field.

The electric field inside the shell is zero due to symmetry and cancellation of field contributions from different parts of the shell.

Symmetry simplifies the calculation by allowing us to deduce the direction of the electric field and reduce the complexity of the integral.

Non-uniform charge distribution complicates the calculation, requiring integration with a charge density that varies with position.