Question

A charge $17.7 \times 10^{-4}\text{ C}$ is distributed uniformly over a large sheet of area $200 \text{ m}^2$. The electric field intensity at a distance $20 \text{ cm}$ from it in air will be $[\epsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{Nm}^2]$}

• A. $5 \times 10^5 \text{ N/C}$
• B. $6 \times 10^5 \text{ N/C}$
• C. $7 \times 10^5 \text{ N/C}$
• D. $8 \times 10^5 \text{ N/C}$

Hint:

For a large (infinite) plane sheet of charge, the electric field intensity $E = \frac{\sigma}{2\epsilon_0}$ is uniform and does not depend on the distance from the sheet. Always check if given information is relevant to the formula used.

Answer 1

The Correct Option is C

Concept:
Electric field due to a large uniformly charged plane sheet is: \[ E=\frac{\sigma}{2\varepsilon_0} \] where:

• $\sigma$ = surface charge density
• $\varepsilon_0$ = permittivity of free space

For an infinite plane sheet, electric field is constant and does not depend on distance from the sheet.
Step 1: Find surface charge density
Given: \[ Q=17.7\times 10^{-4}\ C \] \[ A=200\ m^2 \] Surface charge density: \[ \sigma=\frac{Q}{A} \] \[ \sigma=\frac{17.7\times10^{-4{200} \] \[ \sigma=8.85\times10^{-6}\ C/m^2 \]
Step 2: Use electric field formula
\[ E=\frac{\sigma}{2\varepsilon_0} \] Given: \[ \varepsilon_0=8.85\times10^{-12}\ C^2N^{-1}m^{-2} \] Substitute: \[ E=\frac{8.85\times10^{-6{2(8.85\times10^{-12})} \]
Step 3: Simplify
\[ E=\frac{10^{-6{2\times10^{-12 \] \[ E=\frac{1}{2}\times10^6 \] \[ E=5\times10^5\ N/C \]
Step 4: Distance information
The given distance $20\,cm$ is not used because field due to an infinite sheet remains same everywhere.
Step 5: Final Answer
\[ \boxed{5\times10^5\ N/C} \] Quick Tip:
For plane sheet charge, electric field is uniform and independent of distance.