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Chapter 22: Problem 12

A solid sphere contains positive charge uniformly distributed throughout its volume. Is the potential at its center higher than, lower than, or the same as at the surface?

Short Answer

Expert verified
The electrical potential at the center of the solid sphere is lower than the potential at its surface.

Step by step solution

01

State the problem

We aim to know whether the electric potential at the center of a uniformly charged solid sphere is higher, lower or the same compared to its surface.
02

Recall the definition of electric potential

Electric potential at a point in a field is defined as the work done in carrying a unit positive charge from infinity to that point. Mathematically, this is expressed as \( V = \frac{kq}{r} \) where \( V \) is the electric potential, \( k \) is Coulomb's constant, \( q \) is the total charge, and \( r \) is the distance from the center to the point of interest.
03

Discuss the electric potential at the surface and at the center of the sphere

In the case of a uniformly charged sphere, all the charge can be considered to reside at the center for the points outside and on the surface of the sphere. Thus, at the surface, the electric potential is \( V_s = \frac{kq}{R} \) where \( R \) is the radius of the sphere. However, the charge enclosed within a sphere of radius \( r \) (where \( r<R \)) containing the center point would only be a fraction of the total charge \( q \). Thus, the electric potential at the center (\( r=0 \)) is zero. So the potential at its center is lower than the potential at the surface.

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Most popular questions from this chapter

A dipole of moment \(p=2.9 \mathrm{nC} \cdot \mathrm{m}\) consists of two charges separated by far less than \(10 \mathrm{cm} .\) Find the potential \(10 \mathrm{cm}\) from the dipole (a) on its axis, ( \(b\) ) at \(45^{\circ}\) to its axis, and (c) on its perpendicular bisector.

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