Question

Two infinite parallel planes are separatad 022 cm. The planes hane eoqalo opposite charge densitics The charge densiyof the poisire pane 94 Calculate a) the magitude of the eti le env pa 11 M b) the potential difference from the c) the potential along the equipotential surface 010 m from the potive pla Let the potential along the postive plane cqual 3.0kV 1.9

Answer 1

The electric field due to a single infinite plane with uniform charge density $\sigma$ is given by:

$\vec E_+=\frac{\sigma}{2\epsilon}\hat n$

If the plate is positively charged, the electric field is directed away from the plate, else if the plate is negatively charged, the field is directed towards the plate.

The direction of the electric field is normal to the plane.

Hence for the combination of two planes one with +ve and the other with -ve charge

Hence in between the plates, the net electric field is given as

$\vec E_++\vec E_-= =\frac{\sigma}{2\epsilon}\hat n+\frac{\sigma}{2\epsilon}\hat n==\frac{\sigma}{\epsilon}\hat n$

[note : Here we have already considered the charge on the other plate to be negative, by assigning the direction of the electric field towards the plate, hence in the magnitude part of the electric field, we have taken only the magnitude of $\sigma$ ].

Hence the net electric field is

$\\ \vec E_{net}=\frac{\sigma}{\epsilon_o}\hat n= \frac{9.4\times 10^{-6}C/m^2}{8.84\times 10^{-12}C/m \ V}\hat n=1.06\times 10^6 V/m \\ \vec E_{net}\approx 1.1\times MV/m$

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b)

The potential difference is

$\Delta V=-\int_{0}^{d}\vec E_{net}.d\vec x$

d is the distance between the plates.

here dx is in the direction of the electric field (the integration is from the +ve plate to the -ve plate)

$\Delta V=-\int_{0}^{0.22 \ cm}(1.06\times MV/m.)d\vec x$

$\\ \Delta V=-(1.06\times MV/m.)\int_{0}^{0.10 \ cm}dx=(1.06\times MV/m.)[0.22cm-0cm] \\ \Delta V= -2.3kV$

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c)

Here the upper limit of the integration is changed to 0.10cm.

$\\ \Delta V=-(1.06\times MV/m.)\int_{0}^{0.10 \ cm}dx=(1.06\times MV/m.)[0.10cm-0cm] \\ \Delta V= -1.06kV \\ V_{0.1cm}-V_0= -1.06kV \\ V_{0.1cm}=V_0 -1.06kV \\ V_{0.1cm}=3.0kV -1.06kV= 1.94 kV \\ V_{0.1cm} \approx 1.9kV$