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Hint: We know that a conducting hollow sphere will have the entire charge on its outer surface and the electric field intensity inside the conducting sphere will be zero. A conducting hollow sphere will have the entire charge on its outer surface and the electric field intensity inside the conducting sphere will be zero. Electric potential, the amount of work needed to move a unit charge from a reference point to a specific point against an electric field. Typically, the reference point is Earth, although any point beyond the influence of the electric field charge can be used. When a conductor is at equilibrium, the electric field inside it is constrained to be zero. Since the electric field is equal to the rate of change of potential, this implies that the voltage inside a conductor at equilibrium is constrained to be constant at the value it reaches at the surface of the conductor.
Complete step by step answer
We know that since the two spheres are connected by a wire, they will have the same potential
The potential of a sphere is given by $\dfrac{K Q}{R}$
Since the two spheres will have same potential, once the charge is given, they rearrange such that,
$\dfrac{K Q_{1}}{r_{1}}=\dfrac{K Q_{2}}{r_{2}}$
Now we can write that:
$\dfrac{Q_{1}}{r_{1}}=\dfrac{Q_{2}}{r_{2}}$
Hence, we can write that:
$\dfrac{Q_{1}}{Q_{2}}=\dfrac{r_{1}}{r_{2}}$
Now as we know that: $\left(r_{1}>r_{2}\right)$
So, we can write that:
${{Q}_{1}}>{{Q}_{2}}$
Thus, we conclude that:
${{Q}_{A}}>{{Q}_{B}}$
Thus, we can say the charge will be greater on the surface of sphere A.
So, the correct answer is option C.
Note: We know that potential means ability and in physics ability of a system to perform a work. Work done means any object is displaced from one position to another in the presence of force. So, in the electrical system a force which causes movement of charge to some distance in presence of another force is called potential. The electric potential, or voltage, is the difference in potential energy per unit charge between two locations in an electric field. Electrical potential difference is the difference in the amount of potential energy a particle has due to its position between two locations in an electric field. This important concept provides the basis for understanding electric circuits.
Complete step by step answer
We know that since the two spheres are connected by a wire, they will have the same potential
The potential of a sphere is given by $\dfrac{K Q}{R}$
Since the two spheres will have same potential, once the charge is given, they rearrange such that,
$\dfrac{K Q_{1}}{r_{1}}=\dfrac{K Q_{2}}{r_{2}}$
Now we can write that:
$\dfrac{Q_{1}}{r_{1}}=\dfrac{Q_{2}}{r_{2}}$
Hence, we can write that:
$\dfrac{Q_{1}}{Q_{2}}=\dfrac{r_{1}}{r_{2}}$
Now as we know that: $\left(r_{1}>r_{2}\right)$
So, we can write that:
${{Q}_{1}}>{{Q}_{2}}$
Thus, we conclude that:
${{Q}_{A}}>{{Q}_{B}}$
Thus, we can say the charge will be greater on the surface of sphere A.
So, the correct answer is option C.
Note: We know that potential means ability and in physics ability of a system to perform a work. Work done means any object is displaced from one position to another in the presence of force. So, in the electrical system a force which causes movement of charge to some distance in presence of another force is called potential. The electric potential, or voltage, is the difference in potential energy per unit charge between two locations in an electric field. Electrical potential difference is the difference in the amount of potential energy a particle has due to its position between two locations in an electric field. This important concept provides the basis for understanding electric circuits.
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