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Physics Lesson 14.5.4 - Potential Difference between two Conducting Spheres
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Welcome to our Physics lesson on Potential Difference between two Conducting Spheres, this is the fourth 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 Difference between two Conducting Spheres
Let's consider two charged conducting spheres of different potentials V1 and V2 as shown in the figure.
If we connect the two spheres by a conducting wire, an electric field is generated inside the wire as the spheres are at different potentials. Electric charges start flowing from the sphere with the highest potential to that with the lowest potential. This is similar to water when it flows from a higher to a lower level. In this way, the work done by the electric forces causes a charge transfer. This process continues until both spheres reach the same value of potential, i.e. until the potential difference becomes zero. Then, the charges flow stops.
If we take the radii of the above spheres as r1 and r2, it is obvious that they share the total charge in proportion to their radii. After the contact, the spheres will have the charges q1 and q2. From the law of charge conservation, we can write:
When electric potentials of the two spheres become equal, we have
or
Thus, we can write
Therefore, the law of charge distribution is written as
or
Rearranging for q1, we obtain
or
Similarly, for q2 we obtain
Thus, the common potential of each sphere after the contact, is
= k × Q1 + Q2/r1 + r2
The last formula can be generalized for more than two charges as well, i.e.
Example 4
Calculate the work done when a charge of Q1 = 4 μC is moved from the point a to the point b under the effect of a Q2 = 80 μC charge as shown in the figure.
Solution 4
The work done by electric forces is positive. This means the potential energy of the system decreases as both charges are positive. We have r1 = 0.3 and r2 = 0.4 m. Thus,
= Q1 × (Vb - Va )
= Q1 × k × Q2/r2 - k × Q2/r1
= k × Q1 × Q2 × 1/r2 - 1/r1
= 9 × 109 × 4 × 10-6 × 80 × 10-6 × 1/0.4 - 1/0.3
= 36 × 80 × -10/12 × 10-3 J
= -2.4 J
You have reached the end of Physics lesson 14.5.4 Potential Difference between two Conducting Spheres. There are 8 lessons in this physics tutorial covering Electric Potential, you can access all the lessons from this tutorial below.
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