Question

Charge between parallel plates :

Surface charge density is defined as the charge per unit surface area of surface charge distribution. i.e., $\sigma =\frac{q}{S}$

Two large thin metal plates are parallel and close to each other, on their inner faces, the plates have surface charge densities of opposite sign having magnitude of $1.70\times {10}^{-22}{\text{Cm}}^{-2}$ as shown in figure.
[Use ${\epsilon }_0=8.85\times {10}^{-12}{\text{Fm}}^{-1}$]

$\left(ii\right)$The electric field $E$ in the region to the left of plate $\text{A}$ and to the right of plate $\text{B}$ is

• A. $1.2\times {10}^{-10}{\text{NC}}^{-1};1.2\times {10}^{-10}{\text{NC}}^{-1}$
• B. $1.2\times {10}^{-12}{\text{NC}}^{-1};1.2\times {10}^{-12}{\text{NC}}^{-1}$
• C. $0.2\times {10}^{-10}{\text{NC}}^{-1};0.2\times {10}^{-10}{\text{NC}}^{-1}$
• D. $0{\text{NC}}^{-1};0{\text{NC}}^{-1}$

Answer 1

The correct option is D $0{\text{NC}}^{-1};0{\text{NC}}^{-1}$



Given, surface charge densities of plate <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{A}$ and $\text{B}$ is,

${\sigma }_A=+1.70\times {10}^{-22}{\text{Cm}}^{-2};{\sigma }_B=-1.70\times {10}^{-22}{\text{Cm}}^{-2}$

Using gauss's law, electric field outside the region of plate $\text{A}$ is,

$E=E_A-E_B=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0$

Similarly, electric field outside the region of plate $\text{B}$ is,

$E=E_A-E_B=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.