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3 years ago

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How many electrons must be removed from each of two 4.85-kg copper spheres to make the electric force of repulsion between them

equal in magnitude to the gravitational attraction between them?

1 answer:

VladimirAG [237]3 years ago

3 0

Answer:

$n=2.611*10^{9}electrons$

Explanation:

Given data

mass of copper m=4.85 kg

charge of electron qe= -1.6×10⁻¹⁹C

To find

Number of electron n must be removed

Solution

Equate the magnitude of electric force Fe of repulsion between two two spheres to the magnitude of gravitational force of attraction between them

So

$F_{e}=F_{g}\\ k\frac{q_{e}q_{e}}{r^{2} }=G\frac{m^{2} }{r^{2} }\\ k(q_{e})^{2}=Gm^{2}\\$

where q is charge of each sphere which is equal to number n of removed electrons multiplied by each charge qe

So

$k(nq_{e})^{2}=Gm^{2}\\n=\sqrt{\frac{Gm^{2}}{k(q_{e})^{2}} }\\ n=\sqrt{\frac{(6.67*10^{-11}N.m^{2}/kg^{2})(4.85kg)^{2}}{(8.99*10^{9}N.m^{2}/C^{2} )(1.6*10^{-16}C)^{2}} }\\n=2.611*10^{9}electrons$

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The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

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$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

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Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

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