Electric Field Video Lessons

Concept

# Problem: Charges -q and +2q in the figure are located at x=+-a.  Determine the electric field at points 1 to 4. Write each field in component form.

###### FREE Expert Solution

The electric field due to point charge at distance r:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{k}\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}}$, where q is the charge, r is the distance from the point to the charge, and k is the electrostatic constant.

k is also expressed as:

$\overline{){\mathbf{k}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}}$

Therefore, we have the electric field written as:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{r}}}$

We'll consider the following figure.

At point 1

The distance between charge 2q and point 1 is:

$\begin{array}{rcl}{{\mathbf{r}}_{\mathbf{1}}}^{\mathbf{2}}& \mathbf{=}& {\mathbf{\left(}\mathbf{-}\mathbf{a}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\mathbf{2}\mathbf{a}\mathbf{\right)}}^{\mathbf{2}}\\ & \mathbf{=}& {\mathbf{a}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{a}}^{\mathbf{2}}\end{array}$

r12 = 5a2

The distance between charge, -q and point 1 is:

$\begin{array}{rcl}{{\mathbf{r}}_{\mathbf{2}}}^{\mathbf{2}}& \mathbf{=}& {\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\mathbf{2}\mathbf{a}\mathbf{\right)}}^{\mathbf{2}}\\ & \mathbf{=}& {\mathbf{a}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{a}}^{\mathbf{2}}\end{array}$

r22 = 5a2

From our figure:

$\begin{array}{rcl}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }& \mathbf{=}& \frac{\mathbf{a}}{{\mathbf{r}}_{\mathbf{2}}}\\ & \mathbf{=}& \frac{\mathbf{a}}{\sqrt{\mathbf{5}{\mathbf{a}}^{\mathbf{2}}}}\\ & \mathbf{=}& \frac{\overline{)\mathbf{a}}}{\overline{)\mathbf{a}}\sqrt{\mathbf{5}}}\end{array}$

sinθ = 1/sqrt(5)

Also,

$\begin{array}{rcl}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }& \mathbf{=}& \frac{\mathbf{2}\mathbf{a}}{{\mathbf{r}}_{\mathbf{2}}}\\ & \mathbf{=}& \frac{\mathbf{2}\mathbf{a}}{\sqrt{\mathbf{5}{\mathbf{a}}^{\mathbf{2}}}}\\ & \mathbf{=}& \frac{\mathbf{2}\overline{)\mathbf{a}}}{\overline{)\mathbf{a}}\sqrt{\mathbf{5}}}\end{array}$

cosθ = 2/sqrt(5)

The electric field at point 1 due to charge, +2q is:

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###### Problem Details

Charges -q and +2q in the figure are located at x=+-a.  Determine the electric field at points 1 to 4. Write each field in component form.