Two positrons (e^+) and two protons (p) are kept on four corners of a square of side a as shown

Gujarati

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2. Electric Potential and Capacitance

hard

Two positrons $(e^+)$ and two protons $(p)$ are kept on four corners of a square of side $a$ as shown in figure. The mass of proton is much larger than the mass of positron. Let $q$ denotes the charge on the proton as well as the positron then the kinetic energies of one of the positrons and one of the protons respectively after a very long time will be-

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

$\frac{{{q^2}}}{{2\pi { \in _0}a}},\frac{{{q^2}}}{{4\sqrt 2 \pi { \in _0}a}}$

$\frac{{{q^2}}}{{4\pi { \in _0}a}},\frac{{{q^2}}}{{4\pi { \in _0}a}}$

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

Solution

As mass of proton $>>>$ mass of positron so initial acceleration of positron is much larger than proton. Therefore position reach far away in very short time as compare to proton.

$2 \mathrm{K}_{\mathrm{e}^{+}}=\left(\frac{4 \mathrm{kq}^{2}}{\mathrm{a}}+\frac{2 \mathrm{kq}^{2}}{\mathrm{a} \sqrt{2}}\right)-\frac{\mathrm{kq}^{2}}{\mathrm{a} \sqrt{2}}$

$ \Rightarrow {{\rm{K}}_{{{\rm{e}}^ + }}} = \frac{{{{\rm{q}}^2}}}{{2\pi { \in _0}{\rm{a}}}}\left( {1 + \frac{1}{{4\sqrt 2 }}} \right)$ and

$2{{\rm{K}}_{\rm{p}}} = \frac{{{\rm{k}}{{\rm{q}}^2}}}{{{\rm{a}}\sqrt 2 }} – 0 \Rightarrow {{\rm{K}}_{\rm{p}}} = \frac{{{{\rm{q}}^2}}}{{8\sqrt 2 \pi { \in _0}{\rm{a}}}}$

Standard 12

Physics