Question

Two small spheres spaced 20.0 cm apart have equal charge. part a how many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33×10−21 n?

Answer 1

As the each sphere have the same charge, we use the Coulomb law to calculate the charge on the sphere

$F=k\frac{qq}{r^2} \\\\q=\sqrt{\frac{F r^2}{k} }$

Here, $F=3.33\times 10^{-21} N$ and $r=20 cm=20\times10^{-2} m$ and k is constant and its value is  $8.99\times 10^{9} N.m^2/C^2$

Therefore,

$q=\sqrt{\frac{3.33\times10^{-21}\times(20\times10^{-2} m)^2 }{8.99\times 10^{9} N.m^2/C^2 } }\\\\q=1.2\times 10^{-16} C$

Thus, the number of electron present in each sphere is

$N=\frac{q}{e} \\\\ N =\frac{1.2\times 10^{-16} C}{1.6\times10^{-19}C } \\\\ N=0.75\times 10^3=750$

Answer 2

Solution:

Since we have two small spheres:

One Charge = q1 = q  

Force = F = 4.57*10^-21 N  

Other charge = q2 =q  

Distance = r = 20 cm = 0.2 m  

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)  

Using Coulomb's law,  

F=[1/4pieo]q1q2/r^2  

F = [1/4pieo]q^2 / r^2  

q^2 =F [4pieo]r^2  

q = r*sq rt F[4pieo]  

q=0.2* sq rt[3.33 x 10^-21]*[4*3.1416*8.854*10^-12]  

q = 1.28923*10^ -16 C  

So,

number of excess electrons = n = q/e=1.28923*10^ -16 /1.6*10^-19  

n =750  

Thus,

750 excess electrons must be present on each sphere.