What should be the charge on a sphere of radius 2 cm so that when it is brought in contact with another sphere of radius 5 cm carrying a charge of \[10\mu C\], there is no transfer of charge between the spheres?
Answer
583.5k+ views
Hint: The flow of charge happens due to a potential difference between two points. WE will use this fact to determine the charge on the smaller sphere such that both the spheres have the same potential.
Formula used: In this solution, we will use the following formula:
Potential due to a charged sphere on its surface $V = \dfrac{{kq}}{r}$ where $k$ is a constant, $q$ is the charge on the sphere and $r$ is the radius of the sphere
Complete step by step answer:
We know that the flow of charges occurs when there is a potential difference between any two points. The flow of charge or the transfer of charge between two spheres will hence not occur if both the spheres have the same potential. We know that the potential of a sphere is calculated as
$V = \dfrac{{kq}}{r}$
Now, both the spheres need to have the same potential. We can write that the potential due to the first sphere which has a charge $q$ and radius $2\,cm$ is ${V_1} = \dfrac{{kq}}{2}$. Similarly, the potential at the surface of the second sphere will be ${V_2} = \dfrac{{k(10)}}{5}$. Since both these potentials will be equal, we can write
$\dfrac{{kq}}{2} = \dfrac{{k(10)}}{5}$
Solving for $q$, we get
$q = 4\mu C$
Note: Since we are comparing to potentials, our answer to the charge on the smaller sphere will have the same units as that of the larger sphere. Here the formula that we have used is the potential due to a charged sphere on its surface. Since both the spheres are touching, the potential on the surface of the charged spheres should be the same to avoid the flow of charges.
Formula used: In this solution, we will use the following formula:
Potential due to a charged sphere on its surface $V = \dfrac{{kq}}{r}$ where $k$ is a constant, $q$ is the charge on the sphere and $r$ is the radius of the sphere
Complete step by step answer:
We know that the flow of charges occurs when there is a potential difference between any two points. The flow of charge or the transfer of charge between two spheres will hence not occur if both the spheres have the same potential. We know that the potential of a sphere is calculated as
$V = \dfrac{{kq}}{r}$
Now, both the spheres need to have the same potential. We can write that the potential due to the first sphere which has a charge $q$ and radius $2\,cm$ is ${V_1} = \dfrac{{kq}}{2}$. Similarly, the potential at the surface of the second sphere will be ${V_2} = \dfrac{{k(10)}}{5}$. Since both these potentials will be equal, we can write
$\dfrac{{kq}}{2} = \dfrac{{k(10)}}{5}$
Solving for $q$, we get
$q = 4\mu C$
Note: Since we are comparing to potentials, our answer to the charge on the smaller sphere will have the same units as that of the larger sphere. Here the formula that we have used is the potential due to a charged sphere on its surface. Since both the spheres are touching, the potential on the surface of the charged spheres should be the same to avoid the flow of charges.
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