JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 21)
A system of three charges are placed as shown
in the figure :
_9th_April_Morning_Slot_en_21_1.png)
If D >> d, the potential energy of the system is
best given by :
If D >> d, the potential energy of the system is
best given by :$${1 \over {4\pi {\varepsilon _0}}}\left[ { {{{q^2}} \over d} + {{qQd} \over {{D^2}}}} \right]$$
$${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} - {{qQd} \over {2{D^2}}}} \right]$$
$${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} - {{qQd} \over {{D^2}}}} \right]$$
$${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} + {2{qQd} \over {{D^2}}}} \right]$$
Explanation
Utotal = Uself of dipole + Uinteraction
= $$ - {{k{q^2}} \over d} - \left( {{{kQ} \over {{D^2}}}} \right)qd$$
$$ = - k\left[ {{{{q^2}} \over d} + {{qQd} \over {{D^2}}}} \right]$$
= $$ - {{k{q^2}} \over d} - \left( {{{kQ} \over {{D^2}}}} \right)qd$$
$$ = - k\left[ {{{{q^2}} \over d} + {{qQd} \over {{D^2}}}} \right]$$
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