two small spheres spaced 20.0cm apart have equal

Answered question

2022-01-19

two small spheres spaced 20.0cm apart have equal charges. How many extra electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33*10-21 N

Answer & Explanation

xleb123

xleb123

Skilled2022-01-29Added 181 answers

We are given two charged spheres with the same charge qqq separated by a distance. r=20.0 cm. 
The spheres exerted a force on each other where the magnitude of this force is great F=3.33×1021 N. We are asked to calculate the excess electrons on each sphere.

The number of electrons could be obtained by knowing the charge q of the sphere, so we can use Coulomb's law to obtain the magnitude of the charge q in the form

(q1=q2)F=kq1q2r2(Solve for q)F=kq2r2(1)q=Fr2k

Where k is Coulomb constant and equals 9.0×109 Nm2/C2

Now we can plug our values for F,r and k into equation (1) to get q

q =Fr2k  =(3.33×1021 N)(20.0×102 m)29.0×109 Nm2/C2  =1.22×1016 C

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

(2)ne=qe

Where eee is the charge of the electron and equals 1.602×1019 C/electron. Now we can plug our values for eee and qqq into equation (2) to get the excess of electrons for one sphere

ne =qe=1.22×1016 C1.602×1019 C/electron=760 electron

Each sphere has 760 excess of electrons

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