Question

A and B are two hollow metal spheres of radii 50 cm and 1 m carrying charges 0.6 μC and 1 μC respectively. They are connected externally by a conducting wire. Now the charge flows from:
A. A to B till the charges become equal
B. A to B till the potentials become equal
C. B to A till the charges become equal
D. B to A till the potentials become equal

Answer 1

Hint: The object gains some amount of energy when moved against the electric field that it is defined as the electric potential energy.

Formula used:
${\text{V = }}\dfrac{{{\text{kQ}}}}{{\text{r}}}$

Complete step by step answer:
A hollow metal sphere with uniform charge behaves as if the entire charge were concentrated at its center.
Also, as the electric field is zero inside the metal sphere the electric potential is the same from surface to center.
Given that A and B are two hollow metal spheres that have radii 50 cm and 1 m carrying charges 0.6 μC and 1 μC respectively.
Let ${{\text{r}}_{\text{A}}}$ be the radius of metal sphere A, ${{\text{r}}_{\text{B}}}$ be the radius of metal sphere B, ${{\text{Q}}_{\text{A}}}$be the charge carrying by metal A and ${{\text{Q}}_{\text{B}}}$be the charge carrying by metal B.
Using the equation for the electric potential due to a point charge given by the formula ${\text{V = }}\dfrac{{{\text{kQ}}}}{{\text{r}}}$
Where V is the electric potential, k is the constant$ = 9.0 \times 10_{}^9{\text{ }}\dfrac{{{\text{N - m}}_{}^{\text{2}}}}{{{\text{C}}_{}^2}}$ , Q is the charge and r is the radius of the metal sphere.
Let ${{\text{V}}_{\text{A}}}$ be the electric potential due to sphere A and ${{\text{V}}_{\text{B}}}$ be the electric potential due to sphere B.
Given that \[{{\text{r}}_{\text{A}}}{\text{ = 50 cm = }}50 \times 10_{}^{ - 2}{\text{ m, }}{{\text{r}}_{\text{B}}}{\text{ = 1m, }}{{\text{Q}}_{\text{A}}}{\text{ = 0}}{\text{.6 }}\mu {\text{C = 0}}{\text{.6 }} \times 10_{}^{ - 6}{\text{ C, }}{{\text{Q}}_{\text{B}}}{\text{ = 1}}\mu {\text{C = 1}} \times 10_{}^{ - 6}{\text{ C }}\]
Substituting the values for metal sphere A we get,
\[
   \Rightarrow {{\text{V}}_{\text{A}}} = \dfrac{{9 \times 10_{}^9 \times 0.6 \times 10_{}^{ - 6}}}{{50 \times 10_{}^{ - 2}}} \\
   \Rightarrow {{\text{V}}_{\text{A}}} = 10.8 \times 10_{}^3{\text{ V}} \\
   \Rightarrow {{\text{V}}_{\text{A}}} = 10.8{\text{ kV}} \\
 \]
Similarly, substituting the values for metal sphere B we get,
\[
   \Rightarrow {{\text{V}}_{\text{B}}} = \dfrac{{9 \times 10_{}^9 \times 1 \times 10_{}^{ - 6}}}{1} \\
   \Rightarrow {{\text{V}}_{\text{B}}} = 9 \times 10_{}^3{\text{ V}} \\
   \Rightarrow {{\text{V}}_{\text{B}}} = 9{\text{ kV}} \\
 \]
Hence, ${{\text{V}}_{\text{A}}} \gg {{\text{V}}_{\text{B}}}$
When metal sphere A and B are connected by a wire, charges will flow until their potentials become equal.
The charges that flow through the wire are free electrons and when two bodies at different electric potentials are connected by a metallic wire, electrons flow from the body at lower potential to the one at the higher potential.
Hence, the charge will flow from metal sphere B to A until the potential becomes equal.
Therefore (D) B to A till the potentials become equal is the required answer.

Note: When two bodies at different electric potentials are connected by a metallic wire, electrons flow from the body at lower potential to the one at the higher potential.
But the conventional current will flow from the body at higher potential to the body at lower potential.