Question

Three infinite plane sheets carrying uniform charge densities $-\sigma,\,2\sigma,\,3\sigma$ are placed parallel to the $xz$-plane at $y=a,\;3a,\;4a$, respectively. The electric field at the point $(0,2a,0)$ is

• A. $\dfrac{4\sigma}{\varepsilon_{0}}\,\hat{y}$
• B. $\dfrac{3\sigma}{\varepsilon_{0}}\,\hat{y}$
• C. $\dfrac{2\sigma}{\varepsilon_{0}}\,\hat{y}$
• D. $\dfrac{\sigma}{\varepsilon_{0}}\,\hat{y}$

Hint:

For infinite sheets, direction depends only on whether the point is above or below the sheet—not on distance.

Answer 1

The Correct Option is B

Step 1: Recall field of an infinite sheet.
An infinite plane with charge density $\sigma$ produces electric field
$E = \dfrac{\sigma}{2\varepsilon_0}$
directed away from the positive sheet and toward the negative sheet.

Step 2: Locate the point relative to the sheets.
Point is at $y=2a$.
Sheets are at $y=a$ (negative), $y=3a$ (positive $2\sigma$), $y=4a$ (positive $3\sigma$).

Step 3: Determine direction of each field.
- For sheet at $y=a$ (negative charge): field at $2a$ points toward the sheet ⇒ downward (−$\hat{y}$). Magnitude $=\sigma/(2\varepsilon_0)$.
- For sheet at $y=3a$ (positive $2\sigma$): point is below the sheet ⇒ field downward (−$\hat{y}$). Magnitude $=2\sigma/(2\varepsilon_0)=\sigma/\varepsilon_0$.
- For sheet at $y=4a$ (positive $3\sigma$): point is below the sheet ⇒ field downward (−$\hat{y}$). Magnitude $=3\sigma/(2\varepsilon_0)$.

Step 4: Add magnitudes (all downward).
Total field:
$E = \left(\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{\varepsilon_0}+\frac{3\sigma}{2\varepsilon_0}\right)$ downward.
$E = \frac{4\sigma}{\varepsilon_0}$ downward = $+\dfrac{4\sigma}{\varepsilon_{0}}\hat{y}$ (since downward is +$\hat{y}$ in their diagram).

Step 5: Conclusion.
The electric field at $(0,2a,0)$ is $\dfrac{4\sigma}{\varepsilon_0}\,\hat{y}$.