Join WhatsApp Icon JEE WhatsApp Group
Question 44

Two small spherical balls of mass 10g each with charges $$-2\mu C$$ and $$2\mu C$$, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite non-conducting charge sheet with uniform charge density $$100 \mu C/m^2$$ such that length of rod makes an angle of 30° with electric field generated by charge sheet. Net torque acting on the rod is: (Take $$\varepsilon_0 : 8.85 \times 10^{-12} C^2/Nm^2$$)

Electric field due to infinite non-conducting sheet is

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

Given

$$\sigma=100\mu C/m^2=10^{-4}C/m^2$$

So

$$E=\frac{10^{-4}}{2(8.85\times10^{-12})}$$

$$=5.65\times10^6\text{ N/C}$$

Force on each charge:

F=qEF

with

$$q=2\times10^{-6}C$$

So

$$F=(2\times10^{-6})(5.65\times10^6)$$

=11.3N

The two equal and opposite charges form an electric dipole.

Dipole moment

p=qd

where

d=20cm=0.2m

So

$$p=(2\times10^{-6})(0.2)=4\times10^{-7}$$

Torque on dipole in electric field:

$$\tau=pE\sin\theta$$

with

$$\theta=30^{\circ}$$

Thus

$$\tau=(4\times10^{-7})(5.65\times10^6)\cdot\frac{1}{2}​$$

$$=1.13Nm$$

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.