91影视

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

Use Gauss鈥� 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 鈥渋nside鈥� region between the plates, the individual fields are in the same direction $\left(-\hat{i}\right)$.

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

Here, ${蔚}_0$is the permittivity of free space having a value $8.85脳{10}^{-12}{C}^2/N路m^2$.

Substitute known values in the above equation.

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

In the 鈥渙utside鈥� region where x > 0.5 m, the individual fields point in opposite directions.

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

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

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

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