COMEDK 2025 Evening Shift | Electrostatics Question 3 | Physics

A particle of mass 3 g and charge $60 \mu \mathrm{C}$ is released from rest in a uniform electric field of intensity $10^5 N C^{-1}$. If the value of kinetic energy attained by the particle after moving through a distance of 2 cm is $m \times 10^{-2} J$, then the value of m is:

COMEDK 2025 Evening Shift

MCQ (Single Correct Answer)

-0

A uniformly charged conducting sphere of 0.2 m diameter has a surface charge density of $70 \mu \mathrm{Cm}^{-2}$. The electric flux leaving the surface of the sphere is:

$9.9 \times 10^5 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

$9.9 \times 10^6 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

$8.9 \times 10^5 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

$8.9 \times 10^6 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

COMEDK 2025 Evening Shift

MCQ (Single Correct Answer)

-0

Two electric dipoles of dipole moments $3.9 \times 10^{-30} \mathrm{Cm}$ and $5.2 \times 10^{-30} \mathrm{Cm}$ are placed in two different uniform electric fields of strengths $16 \times 10^4 \mathrm{NC}^{-1}$ and $4 \times 10^4 \mathrm{NC}^{-1}$ respectively. What is the ratio of maximum torque experienced by the electric dipoles?

$12: 1$

$9: 1$

$1:9$

$3: 1$

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

-0

A particle ' X ' carrying a charge +Q is moving in a circular path of radius R around another particle ' Y ' having a charge -Q with a frequency ' $v$ '. Then the mass ' $m$ ' of the charged particle is

$M=\frac{Q^2}{\left[16 \pi^3 \varepsilon_0 R^3 v^2\right]}$

$M=\frac{Q^2}{\left[16 \pi^2 \varepsilon_0 R v^2\right]}$

$M=\frac{Q^2}{\left[16 \pi^2 \varepsilon_0 R^3 v^2\right]}$

$M=\frac{Q^2}{\left[16 \pi^3 \varepsilon_0 R^2 v^2\right]}$

PREVIOUS NEXT

COMEDK Subjects