Two particles each of mass m and charge q are | vedclass

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

Question

(D) Due to symmetry,each particle will have the same speed $v$.The initial total energy of the system is $E_i = \frac{1}{4\pi \varepsilon_0} \frac{q^2

Vedclass · Accepted Answer

none of these

Vedclass · Answer

(D) Due to symmetry,each particle will have the same speed $v$.The initial total energy of the system is $E_i = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{r_1}$.The final total energy of the system at distance $r_2$ is $E_f = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{r_2} + \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{r_2} + mv^2$.Since the electrostatic force is conservative,we apply the law of conservation of energy: $E_i = E_f$.$\frac{1}{4\pi \varepsilon_0} \frac{q^2}{r_1} = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{r_2} + mv^2$.Rearranging for $mv^2$: $mv^2 = \frac{q^2}{4\pi \varepsilon_0} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) = \frac{q^2}{4\pi \varepsilon_0} \left( \frac{r_2 - r_1}{r_1 r_2} \right)$.Therefore,$v = q \sqrt{\frac{r_2 - r_1}{4\pi \varepsilon_0 m r_1 r_2}}$.

Two particles each of mass $m$ and charge $q$ are separated by distance $r_1$ and the system is left free to move at $t = 0$. At time $t$,both the particles are found to be separated by distance $r_2$. The speed of each particle is

Similar Questions

The adjacent diagram shows a charge $+Q$ held on an insulating support $S$ and enclosed by a hollow spherical conductor. $O$ represents the centre of the spherical conductor,and $P$ is a point such that $OP = x$ and $SP = r$. The electric field at point $P$ will be

Two identical charged spheres of material density $\rho$,suspended from the same point by inextensible strings of equal length,make an angle $\theta$ between the strings. When suspended in a liquid of density $\sigma$,the angle $\theta$ remains the same. The dielectric constant $K$ of the liquid is

Find the force experienced by the semicircular rod charged with a charge $q$,placed as shown in the figure. The radius of the wire is $R$ and the line of charge with linear charge density $\lambda$ is passing through its center and is perpendicular to the plane of the wire.

Vedclass Test Series

Exam Paper Generator

Online Exam Module