Question

Two plates in parallel have charge density of area (+)= 10 ? ?∕?2 and ?(−)= 5 ? ?∕?2 , for positive and negative charges respectively. Determine magnitude of electric field on A, B and C areas! (assume ε_0 = 8.85×10−12 ?2∕?.?2)

Answer 1

Given the surface charge densities of the two paltes as

o(+) = 104C/m3

$\sigma (-) = 5\mu C/m^{2}$

$\varepsilon _{0} = 8.85\times 10^{-12}C^{2}/Nm^{2}$

The electric field due to a plane sheet of charge

E 250

Let the two plates be placed as shown in the figure below.

Đ Itt + + + (+) BE C

The three regions were marked as A, B and C. Then the resultant field in region A where the field due to the +ve plate is towards the -ve x direction while that due to the -ve plate is along the +x axis is given by

$E_{A} = -\frac{\sigma (+)}{2\varepsilon _{0} } +\frac{\sigma (-)}{2\varepsilon _{0} }$

  $= -\frac{10\times 10^{-6}}{2\times 8.85\times 10^{-12}} +\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}$

  $= -2.82\times 10^{5} NC^{-1}$

The field in between the two plates is given by

$E_{B} = \frac{\sigma (+)}{2\varepsilon _{0} } +\frac{\sigma (-)}{2\varepsilon _{0} }$

  $= \frac{10\times 10^{-6}}{2\times 8.85\times 10^{-12}} +\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}$

  $= 8.47\times 10^{5} NC^{-1}$

The field in region C is given by

$E_{B} = \frac{\sigma (+)}{2\varepsilon _{0} } -\frac{\sigma (-)}{2\varepsilon _{0} }$

  $= \frac{10\times 10^{-6}}{2\times 8.85\times 10^{-12}} -\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}$

  $= 2.82\times 10^{5} NC^{-1}$