
A hollow metal sphere of radius $10cm$ is charged such that the potential on its surface becomes $80V$. The potential at the centre of the sphere is?
A. $80V$
B. $800V$
C. $8V$
D. $zero$
Answer
583.5k+ views
Hint- Potential at a point is the possibility of charge present at that point. Electric potential at that point is the amount of work done for the unit positive charge \[q\] is brought from infinity to that point.
Formula used: Usually we find the potential at a point due to the charge $q$ by the formula,
$V = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{r}$
Where $r$ is the distance between the charge and the point and $q$ is the charge.
Complete step by step answer:
Consider a charged hollow metal sphere of radius $10cm$. The potential on the surface of that hollow sphere is $80V$. But here we don’t have any need to do this. As it is a hollow sphere, the potential at all the points in the sphere is the same. Therefore, the potential at the centre of the sphere is $80V$.
Electric potential at a point:
The work done in moving a charge of 1 coulomb from infinity to a particular point due to the electric field that is against the electrostatic force then is said to be as the electric potential at a point. And mathematically it is given as
$V = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{r}$
Consider a charged hollow metal sphere of radius $10cm$. The potential on the surface of that hollow sphere is $80V$. But here we don’t have any need to do this. As it is a hollow sphere, the potential at all the points in the sphere is the same. At the centre of the sphere the potential is therefore calculated as $80V$.
Hence the correct option is option A.
Additional information:
(i)Electric potential at P is defined as the amount of work done to move that charge from the infinity to the point P with constant velocity. If the charge is positive, electric potential is greater than zero. If the charge is negative, then the electric potential is negative.
(ii)At infinity, electric potential tends to zero. Because in that case $r = \infty $, therefore electric potential becomes zero as it is inversely proportional to distance $r$.
Note: If the potential at all the points on the surface is equal, then it is called an equipotential surface.
Formula used: Usually we find the potential at a point due to the charge $q$ by the formula,
$V = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{r}$
Where $r$ is the distance between the charge and the point and $q$ is the charge.
Complete step by step answer:
Consider a charged hollow metal sphere of radius $10cm$. The potential on the surface of that hollow sphere is $80V$. But here we don’t have any need to do this. As it is a hollow sphere, the potential at all the points in the sphere is the same. Therefore, the potential at the centre of the sphere is $80V$.
Electric potential at a point:
The work done in moving a charge of 1 coulomb from infinity to a particular point due to the electric field that is against the electrostatic force then is said to be as the electric potential at a point. And mathematically it is given as
$V = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{r}$
Consider a charged hollow metal sphere of radius $10cm$. The potential on the surface of that hollow sphere is $80V$. But here we don’t have any need to do this. As it is a hollow sphere, the potential at all the points in the sphere is the same. At the centre of the sphere the potential is therefore calculated as $80V$.
Hence the correct option is option A.
Additional information:
(i)Electric potential at P is defined as the amount of work done to move that charge from the infinity to the point P with constant velocity. If the charge is positive, electric potential is greater than zero. If the charge is negative, then the electric potential is negative.
(ii)At infinity, electric potential tends to zero. Because in that case $r = \infty $, therefore electric potential becomes zero as it is inversely proportional to distance $r$.
Note: If the potential at all the points on the surface is equal, then it is called an equipotential surface.
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