A particle of mass m and carrying charge -q_1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The electrostatic force between the charges provides the necessary centripetal force for circular motion:$\frac{m v^2}{r} = \frac{1}{4 \pi \vareps

Vedclass · Accepted Answer

$\sqrt{\frac{16 \pi^3 \varepsilon_0 m r^3}{q_1 q_2}}$

Vedclass · Answer

(A) The electrostatic force between the charges provides the necessary centripetal force for circular motion:$\frac{m v^2}{r} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}$Solving for velocity $v$:$v^2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{m r}$$v = \sqrt{\frac{q_1 q_2}{4 \pi \varepsilon_0 m r}}$The period of revolution $T$ is given by the circumference divided by the velocity:$T = \frac{2 \pi r}{v} = 2 \pi r \sqrt{\frac{4 \pi \varepsilon_0 m r}{q_1 q_2}}$$T = \sqrt{(2 \pi r)^2 \cdot \frac{4 \pi \varepsilon_0 m r}{q_1 q_2}}$$T = \sqrt{4 \pi^2 r^2 \cdot \frac{4 \pi \varepsilon_0 m r}{q_1 q_2}}$$T = \sqrt{\frac{16 \pi^3 \varepsilon_0 m r^3}{q_1 q_2}}$

$A$ particle of mass $m$ and carrying charge $-q_1$ is moving around a charge $+q_2$ along a circular path of radius $r$. Find the period of revolution of the charge $-q_1$.

Similar Questions

Two equal negative charges $-q$ are fixed at the points $(0, a)$ and $(0, -a)$ on the $y-$ axis. $A$ positive charge $Q$ is released from rest at the point $(2a, 0)$ on the $x-$ axis. The charge will

An oil drop of radius $2 \, mm$ with a density $3 \, g \, cm^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5} \, V \, m^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (Consider $g = 9.81 \, m \, s^{-2}$)

The intensity of the electric field required to balance a proton of mass $1.7 \times 10^{-27} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is nearly:

An electron and a proton are in a uniform electric field. The ratio of their accelerations will be

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,V/m$. At equilibrium,the angle that the pendulum makes with the vertical is (take $g = 10\,m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module