JEE Main 2024 (Online) 30th January Evening Shift | Electrostatics Question 85 | Physics

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 :

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

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

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

JEE Main 2024 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

An electric field is given by $$(6 \hat{i}+5 \hat{j}+3 \hat{k}) \mathrm{N} / \mathrm{C}$$. The electric flux through a surface area $$30 \hat{i} \mathrm{~m}^2$$ lying in YZ-plane (in SI unit) is :

180

150

JEE Main 2024 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

Two charges of $$5 Q$$ and $$-2 Q$$ are situated at the points $$(3 a, 0)$$ and $$(-5 a, 0)$$ respectively. The electric flux through a sphere of radius '$$4 a$$' having center at origin is :

$$\frac{2 Q}{\varepsilon_0}$$

$$\frac{7 Q}{\varepsilon_0}$$

$$\frac{3 Q}{\varepsilon_0}$$

$$\frac{5 Q}{\varepsilon_0}$$

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