Respuesta :
Answer:
(a). The total charge of this system is 10 C.
(b). The charges of the spheres before the connection is 5 C and 4.2 C.
Explanation:
Given that,
Distance [tex]r= 1.00\times10^{5}\ m[/tex]
Force = 22.5 N
The two spheres are connected for a moment by a wire. After this, charge is re-distributed equally, and each sphere now has a charge
We need to calculate the charge
Using formula of electric force
[tex]F=\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Here, charge is equal
Put the value into the formula
[tex]22.5=\dfrac{9\times10^{9}\times q^2}{(1.00\times10^{5})}[/tex]
[tex]q^2=\dfrac{22.5\times(1.00\times10^{5})^2}{9\times10^{9}}[/tex]
[tex]q=\sqrt{\dfrac{22.5\times(1.00\times10^{5})^2}{9\times10^{9}}}[/tex]
[tex]q=5\ C[/tex]
(a). We need to calculate the total charge of this system
We have the charge of each sphere after re-distribution.
We know that the charge is distributed equally.
Therefore, the total charge of the system is
[tex]q+Q=2q=2\times5=10\ C[/tex]
(b). We need to calculate the charge of the spheres before the connection
Using formula of electric force
[tex]F=\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]18.9=\dfrac{9\times10^{9}\times q_{1}\times q_{2}}{(1.00\times10^{5})}[/tex]
[tex]q_{2}=\dfrac{18.9\times(1.00\times10^{5})^2}{9\times10^{9}\times5}[/tex]
[tex]q_{2}=4.2\ C[/tex]
Hence, (a). The total charge of this system is 10 C.
(b). The charges of the spheres before the connection is 5 C and 4.2 C.
