JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 25)
An infinite number of point charges, each carrying 1 $$\mu$$C charge, are placed along the y-axis at y = 1 m, 2 m, 4 m, 8 m ...............
The total force on a 1C point charge, placed at the origin, is x $$\times$$ 103 N.
The value of x, to the nearest integer, is __________. [Take $${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}$$ Nm2/C2]
The total force on a 1C point charge, placed at the origin, is x $$\times$$ 103 N.
The value of x, to the nearest integer, is __________. [Take $${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}$$ Nm2/C2]
Answer
12
Explanation
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$${F_{total}} = {{k{q_1}{q_2}} \over {r_1^2}} + {{k{q_1}{q_3}} \over {r_2^2}} + {{k{q_1}{q_4}} \over {r_3^2}} + .....$$
$$ = 9 \times {10^9} \times {10^{ - 6}}\left[ {1 + {{\left( {{1 \over 2}} \right)}^2} + {{\left( {{1 \over {{2^2}}}} \right)}^2} + {{\left( {{1 \over {{2^3}}}} \right)}^2} + {{\left( {{1 \over {{2^\infty }}}} \right)}^2}} \right]$$
$$ = 9 \times {10^9} \times {10^{ - 6}}\left[ {{1 \over {1 - {1 \over 4}}}} \right]$$
$$ \because $$ $$\left[ {{S_\infty } = {a \over {1 - r}}} \right]$$ for G.P.
$$ = 9 \times {10^3} \times {4 \over 3}$$ = 12 $$\times$$ 103 N
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