JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 24)
Two identical conducting spheres with negligible volume have 2.1 nC and $$-$$0.1 nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is __________ $$\times$$ 10$$-$$9 N.
[Given : $$4\pi {\varepsilon _0} = {1 \over {9 \times {{10}^9}}}$$ SI unit]
[Given : $$4\pi {\varepsilon _0} = {1 \over {9 \times {{10}^9}}}$$ SI unit]
Answer
36
Explanation
_25th_February_Evening_Shift_en_24_1.png)
When they brought into contact & then separated by a distance = 0.5 m
Then charge distribution will be
_25th_February_Evening_Shift_en_24_2.png)
The electrostatic force acting b/w the sphere is
$${F_e} = {{k{q_1}{q_2}} \over {{r^2}}}$$
$$ = {{9 \times {{10}^9} \times 1 \times {{10}^{ - 9}} \times 1 \times {{10}^{ - 9}}} \over {{{(0.5)}^2}}}$$
$$ = {{900} \over {25}} \times {10^{ - 9}}$$
$$ \Rightarrow $$ $${F_e} = 36 \times {10^{ - 9}}N$$
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