Two uniformly charged, infinite, nonconducting planes are parallel to a yz plane and positioned at x = -50cmand x =+ 50cm. The charge densities on the planes are -50 nC/m²and +25 nC/m² , respectively. What is the magnitude of the potential difference between the origin and the point on the x axis at x = +80cm ?

• The charge density on the first plane is ${\sigma }_1=-50nC/m^2$.
• The charge density on the second plane is ${\sigma }_2=+25nC/m^2$.
• The position of first plane is at x = -50 cm.
• The position of second plane is at x = +50 cm.

Use Gauss’ law to solve the problem.

Gauss's law states that the total electric flux from an enclosed surface is equal to the enclosed charge divided by the permittivity.

In the “inside” region between the plates, the individual fields are in the same direction $\left(-\hat{i}\right)$.

$E_{in}=\left(-\frac{{\sigma }_1}{2{\epsilon }_0}-\frac{{\sigma }_2}{2{\epsilon }_0}\right)$

Here, ${\epsilon }_0$is the permittivity of free space having a value $8.85\times {10}^{-12}{C}^2/N·m^2$.

Substitute known values in the above equation.

$\begin{array}{rcl}{E}_{in}& =& -\left(\frac{50\times {10}^{-9}C/m^2}{2\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)}+\frac{25\times {10}^{-9}C/m^2}{2\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)}\right)\hat{i}\\ & =& -\left(2.8\times {10}^{3}N/C+1.4\times {10}^{3}N/C\right)\hat{i}\\ & =& -\left(4.2\times {10}^{3}N/C\right)\hat{i}\end{array}$

In the “outside” region where x > 0.5 m, the individual fields point in opposite directions.

role="math" localid="1662100706837" $\begin{array}{rcl}{E}_{out}& =& \left(-\frac{{\sigma }_1}{2{\epsilon }_0}+\frac{{\sigma }_2}{2{\epsilon }_0}\right)\\ & =& \left(-\frac{50\times {10}^{-9}C/m^2}{2\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)}+\frac{25\times {10}^{-9}C/m^2}{2\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)}\right)\hat{i}\\ & =& \left(-2.8\times {10}^{3}N/C+1.4\times {10}^{3}N/C\right)\hat{i}\\ & =& -\left(1.4\times {10}^{3}N/C\right)\hat{i}\end{array}$

Therefore, by equation of the electric potential difference between two points i and f,

$\begin{array}{rcl}∆V& =& V_f-V_i\\ & =& -\underset{i}{\overset{f}{\int }}E·ds\\ & =& -\underset{0}{\overset{0.8}{\int }}E·ds\\ & =& -\underset{0}{\overset{0.5}{\int }}\left|E_{in}\right|dx-\underset{0.5}{\overset{0.8}{\int }}\left|E_{out}\right|dx\\ ∆V& =& -\left|E_{in}\right|{\left[x\right]}_0^{0.5}-\left|E_{out}\right|{\left[x\right]}_{0.5}^{0.8}\\ & =& -\left(4.2\times {10}^3\right)\left(0.5\right)-\left(1.4\times {10}^3\right)\left(0.3\right)\\ & =& -2.1\times {10}^3-0.4\times {10}^3\\ & =& 2.5\times {10}^{3}V\end{array}$

Hence, the magnitude of the potential difference between the origin and the point is $2.5\times {10}^{3}V$.