Join WhatsApp Icon JEE WhatsApp Group
Question 43

Two infinite identical charged sheets and a charged spherical body of charge density '$$\rho$$' are arranged as shown in figure. Then the correct relation between the electrical fields at A, B, C and D points is:

image

Field due to one infinite positively charged sheet is

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

directed away from the sheet.

There are two identical positive sheets.

At points A and B, which lie between the sheets:

  • Left sheet produces field to the right.
  • Right sheet produces field to the left.

Magnitudes are equal, so they cancel.

So between the sheets only the charged sphere contributes.

Outside a uniformly charged sphere,

$$E\propto\frac{1}{r^2}$$

Point A is closer to the sphere than BBB, hence

$$E_A>E_B$$

Now point C:

It lies outside both sheets.

Both sheet fields point left and add:

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

Sphere field at C also points left, so it adds.

Thus

$$E_C=\frac{\sigma}{\varepsilon_0}+E_{\text{sphere at C}}$$

At point D:

Again sheet contribution is

$$\frac{\sigma}{\varepsilon_0}$$

Sphere field also adds.

But D is closer to the sphere than C, so

$$E_{\text{sphere at D}}>E_{\text{sphere at C}}$$

Therefore

$$E_D>E_C$$

Get AI Help

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Ask AI

Ask our AI anything

AI can make mistakes. Please verify important information.