Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center) and repel each other with an electrostatic force of F1=0.0720 N . A thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed, the spheres still repel, but with a force of F2=0.115 N . The Coulomb force constant is k=1/(4π????0)=8.99×109 N⋅m2/C2 . Using this information, find the initial charge on each sphere, q1 and q2 , if q1 is initially less than q2 .

Answer 1

`q_1=5.64* 10^(-7)\ \text{C}` and
`q_2=2.32* 10^(-6)\ \text{C}`

Step-by-step explanation:

`F_1=0.072\ \text{N}`

`F_2=0.115\ \text{N}`

r = Distance between shells = 40.4 cm

`q_1` and
`q_2` are the charges

`k` = Coulomb constant =
`8.99*10^(9)\ \text{Nm}^2/\text{C}^2`

Force is given by

`F_1=(kq_1q_2)/(r^2)\\\Rightarrow q_1q_2=(F_1r^2)/(k)\\\Rightarrow q_1q_2=(0.072* 0.404^2)/(8.99* 10^(9))\\\Rightarrow q_1q_2=1.307* 10^(-12)\\\Rightarrow q_1=(1.307* 10^(-12))/(q_2)`

`F_2=(kq^2)/(r^2)\\\Rightarrow q=\sqrt{(F_2r^2)/(k)}\\\Rightarrow q=\sqrt{(0.115* 0.404^2)/(8.99* 10^(9))}\\\Rightarrow q=1.44* 10^(-6)\ \text{C}`

`q=(q_1+q_2)/(2)\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2*1.44* 10^(-6)\\\Rightarrow q_1+q_2=2.88* 10^(-6)`

Substituting the above value of
`q_1` we get

`(1.307* 10^(-12))/(q_2)+q_2=2.88* 10^(-6)\\\Rightarrow q_2^2-2.88* 10^(-6)q_2+1.307* 10^(-12)=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4* \:1* \:1.307* 10^(-12)}}{2* \:1}\\\Rightarrow q_2=2.32* 10^(-6), 5.64* 10^(-7)`

`q_1=(1.307* 10^(-12))/(q_2)=(1.307* 10^(-12))/(2.32* 10^(-6))\\\Rightarrow q_1=5.63* 10^(-7)`

`q_1=(1.307* 10^(-12))/(q_2)=(1.307* 10^(-12))/(5.64* 10^(-7))\\\Rightarrow q_1=2.32* 10^(-6)`

Since we know
`q_1<q_2`

`q_1=5.64* 10^(-7)\ \text{C}` and
`q_2=2.32* 10^(-6)\ \text{C}`.