JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 1)
A two point charges 4q and -q are fixed
on the x-axis at x = $$ - {d \over 2}$$
and x = $${d \over 2}$$
respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will :
_4th_September_Morning_Slot_en_1_1.png)
increase by $${{3{q^2}} \over {4\pi {\varepsilon _0}d}}$$
increase by $${{2{q^2}} \over {3\pi {\varepsilon _0}d}}$$
decrease by $${{{q^2}} \over {4\pi {\varepsilon _0}d}}$$
decrease by $${{4{q^2}} \over {3\pi {\varepsilon _0}d}}$$
Explanation
$$\Delta U = {1 \over {4\pi {\varepsilon _0}}}.{{4q.q} \over {(3d/2)}} - {1 \over {4\pi {\varepsilon _0}}}.{{4q.q} \over {(d/2)}}$$
$$ = {{4{q^2}} \over {4\pi {\varepsilon _0}}}\left( {{2 \over d}} \right)\left( { - {2 \over 3}} \right)$$
= decrease by $${{4{q^2}} \over {3\pi {\varepsilon _0}d}}$$
$$ = {{4{q^2}} \over {4\pi {\varepsilon _0}}}\left( {{2 \over d}} \right)\left( { - {2 \over 3}} \right)$$
= decrease by $${{4{q^2}} \over {3\pi {\varepsilon _0}d}}$$
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