JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 25)
Two electric dipoles, A, B with respective dipole moments $${\overrightarrow d _A} = - 4qai$$ and $${\overrightarrow d _B} = - 2qai$$ are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is -
_10th_January_Morning_Slot_en_25_1.png)
$${{\sqrt 2 R} \over {\sqrt 2 + 1}}$$
$${R \over {\sqrt 2 + 1}}$$
$${{\sqrt 2 R} \over {\sqrt 2 - 1}}$$
$${R \over {\sqrt 2 - 1}}$$
Explanation
V $$ = {{4qa} \over {\left( {R + x} \right)}} = {{2qa} \over {\left( {{x^2}} \right)}}$$
$$\sqrt 2 x = R + x$$
$$x = {R \over {\sqrt 2 - 1}}$$
_10th_January_Morning_Slot_en_25_2.png)
dist $$ = {R \over {\sqrt 2 - 1}} + R = {{\sqrt 2 R} \over {\sqrt 2 - 1}}$$
$$\sqrt 2 x = R + x$$
$$x = {R \over {\sqrt 2 - 1}}$$
dist $$ = {R \over {\sqrt 2 - 1}} + R = {{\sqrt 2 R} \over {\sqrt 2 - 1}}$$
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