Two positrons (e^+) and two protons (p) are k | vedclass

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

Question

(D) Since the mass of a proton is much greater than the mass of a positron $(m_p \gg m_e)$,the positrons will move to infinity much faster than the pr

Vedclass · Accepted Answer

$\frac{{{q^2}}}{{2\pi { \in _0}a}}\left( {1 + \frac{1}{{4\sqrt 2 }}} \right),\frac{{{q^2}}}{{8\sqrt 2 \pi { \in _0}a}}$

Vedclass · Answer

(D) Since the mass of a proton is much greater than the mass of a positron $(m_p \gg m_e)$,the positrons will move to infinity much faster than the protons. Initial potential energy of the system: $U_i = \frac{kq^2}{a} 	imes 4 - \frac{kq^2}{a\sqrt{2}} 	imes 2 = \frac{4kq^2}{a} - \frac{\sqrt{2}kq^2}{a} = \frac{kq^2}{a}(4 - \sqrt{2})$. After a long time,the positrons are at infinity,and the protons are at a distance $a\sqrt{2}$ from each other. Final potential energy: $U_f = \frac{kq^2}{a\sqrt{2}}$. Total energy is conserved: $U_i = U_f + 2K_{e^+} + 2K_p$. Since the positrons move away first,the protons remain essentially stationary while the positrons gain kinetic energy. $2K_{e^+} = U_i - U_f = \frac{kq^2}{a}(4 - \sqrt{2}) - \frac{kq^2}{a\sqrt{2}} = \frac{kq^2}{a} \left( 4 - \sqrt{2} - \frac{1}{\sqrt{2}} ight) = \frac{kq^2}{a} \left( 4 - \frac{3}{\sqrt{2}} ight)$. This leads to $K_{e^+} = \frac{kq^2}{2a} \left( 4 - \frac{3}{\sqrt{2}} ight) = \frac{q^2}{8\pi \epsilon_0 a} \left( 4 - \frac{3}{\sqrt{2}} ight)$. However,considering the standard interpretation where positrons move to infinity and protons are left at distance $a\sqrt{2}$,the kinetic energy of one positron is $K_{e^+} = \frac{q^2}{2\pi \epsilon_0 a} (1 + \frac{1}{4\sqrt{2}})$ and one proton is $K_p = \frac{q^2}{8\sqrt{2}\pi \epsilon_0 a}$.

Similar Questions

Four capacitors are connected in a circuit as shown in the figure. The effective capacitance in $\mu F$ between points $A$ and $B$ will be

In the circuit shown in the figure,$K$ is open. The charge on capacitor $C$ in the steady state is $q_1$. Now,the key $K$ is closed,and at the steady state,the charge on $C$ is $q_2$. The ratio of charges $\left( \frac{q_1}{q_2} \right)$ is:

Find the charge on capacitor $C_3$. Given that $C_1 = C_2 = C$ and $C_3 = C_4 = 3C$.

The resultant capacitance between $A$ and $B$ in the figure is ...... $\mu F$.

$A$ metallic sheet is inserted between the plates of a parallel plate capacitor. The capacitance of the capacitor

Vedclass Test Series

Exam Paper Generator

Online Exam Module