JEE Main 2024 (Online) 31st January Evening Shift | Electrostatics Question 20 | Physics

Force between two point charges $$q_1$$ and $$q_2$$ placed in vacuum at '$$r$$' cm apart is $$F$$. Force between them when placed in a medium having dielectric constant $$K=5$$ at '$$r / 5$$' $$\mathrm{cm}$$ apart will be:

$$5 F$$

$$25 F$$

$$F / 5$$

$$F / 25$$

JEE Main 2024 (Online) 31st January Morning Shift

MCQ (Single Correct Answer)

-1

Two charges $$q$$ and $$3 q$$ are separated by a distance '$$r$$' in air. At a distance $$x$$ from charge $$q$$, the resultant electric field is zero. The value of $$x$$ is :

$$\frac{r}{3(1+\sqrt{3})}$$

$$\frac{(1+\sqrt{3})}{r}$$

$$\frac{r}{(1+\sqrt{3})}$$

$$r(1+\sqrt{3})$$

JEE Main 2024 (Online) 30th January Evening Shift

MCQ (Single Correct Answer)

-1

A particle of charge '$$-q$$' and mass '$$m$$' moves in a circle of radius '$$r$$' around an infinitely long line charge of linear charge density '$$+\lambda$$'. Then time period will be given as :

(Consider $$k$$ as Coulomb's constant)

$$T^2=\frac{4 \pi^2 m}{2 k \lambda q} r^3$$

$$T=\frac{1}{2 \pi r} \sqrt{\frac{m}{2 k \lambda q}}$$

$$T=\frac{1}{2 \pi} \sqrt{\frac{2 k \lambda q}{m}}$$

$$T=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}$$

JEE Main 2024 (Online) 30th January Morning Shift

MCQ (Single Correct Answer)

-1

The electrostatic potential due to an electric dipole at a distance '$$r$$' varies as :

$$\frac{1}{r^3}$$

$$\frac{1}{\mathrm{r}}$$

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