How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 newtons?

Two small spheres spaced 25.0 cm apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 newtons

Answer:

Each sphere has 1112.02 electrons

Explanation:

Force F=4.57×10⁻²¹N

Space r=25 cm=0.25 m

From Coulombs Law we know that

[tex]F=k\frac{q_{1}q_{2} }{r^{2} }\\ Where\\q_{1}=q_{2}=q\\and\\k=9.0*10^{9}N.m^{2}/C^{2}\\ So\\F=k(q^{2}/r^{2} )\\q=\sqrt{\frac{Fr^{2} }{k} } \\q=\sqrt{\frac{(4.57*10^{-21}N )(0.25m)^{2} }{9.0*10^{9}N.m^{2}/C^{2}}[/tex]

q=1.7815×10⁻¹⁶C

After knowing the charge q on sphere we can obtain the number of electrons which is given by:

[tex]n_{e}=\frac{q}{e}\\Where \\e_{electron-charge}=1.602*10^{-19}C/electron \\n_{e}=\frac{(1.7814*10^{-16}C)}{1.602*10^{-19}C/electron}\\n_{e}=1112.02electrons[/tex]

Each sphere has 1112.02 electrons

adioabiola adioabiola · Answer 2 · 2022-01-20T10:02:11+01:00

I'm the complete question; the distance between the two spheres is 25cm

The number of electrons which should be present on each sphere is; 1112.5 electrons.

Coulumb's law

The charge on each electron is; 1.6 × 10-¹⁹ coulombs

Coulumb's constant, q = 8.99×10⁹.

Distance between them = 25cm = 0.25m

Force of repulsion = 4.57×10-²¹

From Coulumb's law;

[tex]f = \frac{k {q}^{2} }{ {r}^{2} }[/tex]

Upon, evaluation;

The value of Q = 1.78 × 10-¹⁶

Therefore, the no. of electrons on each sphere is;

n = q/e

n = 1.78× 10-¹⁶/1.6 × 10-¹⁹

n = 1112.5 electrons