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Physics Lesson 14.5.7 - Potential of a Charged Sphere

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Welcome to our Physics lesson on Potential of a Charged Sphere, this is the seventh lesson of our suite of physics lessons covering the topic of Electric Potential, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Potential of a Charged Sphere

As explained in the previous tutorials, the electric field inside a charged sphere is zero because all charges are distributed uniformly on the surface only. This is because a sphere can naturally have only one type of extra charge, and they repel each other as far as possible as explained earlier.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Potential

Since E = ΔV / d, the potential difference inside the sphere is zero for any d as well. This means the potential inside the sphere is the same everywhere; it is equal to the potential on the outer surface, i.e.

V = k × Q/R

where Q is the amount of charge the sphere carries and R is its radius.

The potential for points outside the sphere is

V = k × Q/r

where r > R is the distance from the centre of sphere to the given point. In this case, the charge contained by the sphere behaves as if it was entirely concentrated at centre of the sphere. With the increase of distance from centre, the potential decreases, as electric potential and distance vary inversely.

Below, a graph expressing the relationship between electric potential and distance from the centre for a charged sphere is shown.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Potential

Example 6

A metal sphere has a charge of Q. What is the relationship between the electric potentials at point K, L, M and N shown in the figure?

Physics Tutorials: This image provides visual information for the physics tutorial Electric Potential

Solution 6

The electric potential at any point inside the sphere and on its surface is

V = k × Q/R

Here, R = 2x. Therefore, the potential at points K and L is

VK = VL = k × Q/2x

The points M and N are equipotential as rM = rN = 3x = 3R/2. Hence,

VM = VN = k × Q/3x

Therefore, we have

VM = VN > VM = VN

Or

VM = VN = 3/2 VM

and

VM = VN = 3/2 VN

You have reached the end of Physics lesson 14.5.7 Potential of a Charged Sphere. There are 8 lessons in this physics tutorial covering Electric Potential, you can access all the lessons from this tutorial below.

More Electric Potential Lessons and Learning Resources

Electrostatics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
14.5Electric Potential
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
14.5.1The Meaning of Electric Potential
14.5.2Potential Difference (Voltage)
14.5.3Analogy between Gravitation and Electricity recap
14.5.4Potential Difference between two Conducting Spheres
14.5.5The Relationship between Potential Difference and Electric Field
14.5.6Equipotential Surfaces
14.5.7Potential of a Charged Sphere
14.5.8Motion of Charged Particles inside a Uniform Field

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  4. Electrostatics Practice Questions: Electric Potential. Test and improve your knowledge of Electric Potential with example questins and answers
  5. Check your calculations for Electrostatics questions with our excellent Electrostatics calculators which contain full equations and calculations clearly displayed line by line. See the Electrostatics Calculators by iCalculator™ below.
  6. Continuing learning electrostatics - read our next physics tutorial: Electric Flux. Gauss Law

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