AIIMS 2006 Physics Electric Fields MCQ Question

To solve the problem of finding the electric field between two parallel large thin metal sheets with equal surface charge densities of opposite signs, we can use the principles of electrostatics.

Explanation:

1. Understanding the Electric Field Due to a Single Charged Sheet: The electric field ($E$) created by an infinite plane sheet of charge with surface charge density $\sigma$ is given by the formula:

   $$E = \frac{\sigma}{2 \varepsilon_0}$$

   where $\varepsilon_0$ is the permittivity of free space, approximately equal to $8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2$.

2. Electric Field Between Two Oppositely Charged Sheets: When two sheets have equal and opposite surface charge densities, the electric fields due to each sheet add up between the sheets and cancel outside. Specifically, if one sheet has a charge density of $\sigma$ and the other has $-\sigma$, the total electric field $E_{\text{between}}$ between the sheets is:

   $$E_{\text{between}} = E_1 + E_2 = \frac{\sigma}{2 \varepsilon_0} + \frac{\sigma}{2 \varepsilon_0} = \frac{\sigma}{\varepsilon_0}$$

3. Calculation of the Electric Field: Given that $\sigma = 26.4 \times 10^{-12} \, \text{C/m}^2$, we can now calculate the electric field between the sheets.

   Using the formula:

   $$E_{\text{between}} = \frac{\sigma}{\varepsilon_0}$$

   Substituting the values:

   $$E_{\text{between}} = \frac{26.4 \times 10^{-12} \, \text{C/m}^2}{8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2}$$

   Performing the calculation:

   $$E_{\text{between}} = \frac{26.4}{8.85} \text{ N/C} \approx 2.99 \text{ N/C}$$

   Rounding gives us approximately $3 \, \text{N/C}$.

4. Conclusion: Therefore, the electric field between the two sheets is approximately $3 \, \text{N/C}$, which corresponds to option C.

Why Other Options Are Incorrect:

• Option A (1.5 N/C): This value is too low, possibly resulting from a misunderstanding of how fields add up between oppositely charged sheets.
• Option B (5 × 10¹⁰ N/C): This value is excessively high and not physically plausible within the context of the given surface charge density.
• Option D (3 × 10¹⁰ N/C): This value is also excessively high, similar to option B, and suggests a significant error in understanding the relationship between charge density and electric field in this scenario.

In summary, the correct answer is C (3 N/C), with the calculation confirming the relationship between charge density and electric field for two parallel plates with equal and opposite charge densities.