
Why is the potential inside a hollow spherical charged conductor, constant and has the same value as on its surface?
Answer
553.8k+ views
Hint:The potential is defined as the work required to move a charge from infinity to a point. Gauss’s theorem states that the total electric flux out of a closed surface is equal to charge enclosed divided by the permittivity. Using the mathematical expression for the potential difference from these theorems, determine the potential inside the hollow spherical charged conductor.
Complete answer:
We know the formula for work done inside a hollow conductor in moving a charge from a point A at infinity to point B is
\[{V_A} - {V_B} = - \int {\vec E \times d\vec l} \]
Here, \[{V_A}\] is the potential at point A, \[{V_B}\] is potential at point B, \[\vec E\] is the electric field and \[d\vec l\] is the change in the length.
As we know that the electric field intensity inside the hollow spherical charged conductor is zero. Hence, the work done in moving a point charge inside the hollow spherical conductor is also zero.
This implies that the potential difference between any two points inside or on the surface of the conductor is zero.
\[ \Rightarrow {V_A} - {V_B} = - \int {\vec E \times d\vec l} = 0\]
\[ \Rightarrow {V_A} - {V_B} = {\text{constant}}\]
From this we can conclude that the potential inside a hollow spherical charged conductor, constant and has the same value as on its surface.
Note:One can also solve the same question by another way. The electric flux in an area is defined as an electric field multiplied by the area of the surface projected in a plane and perpendicular to the field. As per Gauss's theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. We can say that net flux in a closed surface is directly proportional to net charge in volume enclosed by closed surface. Gauss's law is only a result of Coulomb's law. If we apply Gauss’s theorem to a point charge enclosed in a sphere. We get back the coulomb’s law easily. If there are no charges enclosed by a surface, then the net electric flux remains constant.
Complete answer:
We know the formula for work done inside a hollow conductor in moving a charge from a point A at infinity to point B is
\[{V_A} - {V_B} = - \int {\vec E \times d\vec l} \]
Here, \[{V_A}\] is the potential at point A, \[{V_B}\] is potential at point B, \[\vec E\] is the electric field and \[d\vec l\] is the change in the length.
As we know that the electric field intensity inside the hollow spherical charged conductor is zero. Hence, the work done in moving a point charge inside the hollow spherical conductor is also zero.
This implies that the potential difference between any two points inside or on the surface of the conductor is zero.
\[ \Rightarrow {V_A} - {V_B} = - \int {\vec E \times d\vec l} = 0\]
\[ \Rightarrow {V_A} - {V_B} = {\text{constant}}\]
From this we can conclude that the potential inside a hollow spherical charged conductor, constant and has the same value as on its surface.
Note:One can also solve the same question by another way. The electric flux in an area is defined as an electric field multiplied by the area of the surface projected in a plane and perpendicular to the field. As per Gauss's theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. We can say that net flux in a closed surface is directly proportional to net charge in volume enclosed by closed surface. Gauss's law is only a result of Coulomb's law. If we apply Gauss’s theorem to a point charge enclosed in a sphere. We get back the coulomb’s law easily. If there are no charges enclosed by a surface, then the net electric flux remains constant.
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