Question

The infinite sheets of uniform charge density $10\frac{\mu C}{m^2},-10\frac{\mu C}{m^2}and5\frac{\mu C}{m^2}$ are placed in vacuum as shown. The direction of field intensity at points A and B are respectively?

• A. Left, Left
• B. Left, Right
• C. Right, left
• D. Right, Right

Answer 1

The correct option is C

Right, left

Let’s call $E_1,E_2$ and $E_3$, the intensity of fields due to sheet 1,2 and 3 respectively.

At point A:

Contributions from different sheet are as shown –

Where

$|\overrightarrow{E_1}|=\frac{{\sigma }_1}{2{\epsilon }_0}=\frac{10\times {10}^{-6}}{2{\epsilon }_0}=\frac{5\times {10}^{-6}}{{\epsilon }_0}\frac{N}{C}|\overrightarrow{E_2}|=\frac{{\sigma }_2}{2{\epsilon }_0}=\frac{5\times {10}^{-6}}{{\epsilon }_0}\frac{N}{C}|\overrightarrow{E_3}|=\frac{{\sigma }_3}{2{\epsilon }_0}=\frac{5\times {10}^{-6}}{2{\epsilon }_0}\frac{N}{C}|\overrightarrow{E_{nst}}|=|\overrightarrow{E_1}|+|\overrightarrow{E_2}|-|\overrightarrow{E_3}|=\frac{{10}^{-6}}{{\epsilon }_0}5+5-2.5=\frac{7.5\times {10}^{-6}}{{\epsilon }_0}\frac{N}{C},$ towards right

At point B:

Where,

$|\overrightarrow{E_1}|=\frac{5\times {10}^{-6}}{{\epsilon }_0}\frac{N}{C}|\overrightarrow{E_2}|=\frac{5\times {10}^{-6}}{{\epsilon }_0}\frac{N}{C}|\overrightarrow{E_3}|=\frac{5\times {10}^{-6}}{2{\epsilon }_0}\frac{N}{C}|\overrightarrow{E_{nst}}|=|\overrightarrow{E_1}|+|\overrightarrow{E_2}|-|\overrightarrow{E_3}|==\frac{{10}^{-6}}{{\epsilon }_0}\{5-5-\frac{5}{2}\}=\frac{2.5\times {10}^{-6}}{{\epsilon }_0}\frac{N}{C},$

Negative sign shows clearly that this net field at point B is towards left.