A question on electricity (potential difference)

In summary, the conversation discusses the problem of a particle with charge q executing simple harmonic motion in front of a conducting earthed sphere. The goal is to determine the reading of the voltmeter when the particle is at its mean position. The solution involves simplifying the problem to a classical electrostatic one and solving for the potential V of the sphere. The differential equation for V can be simplified if the amplitude A of motion is much smaller than the distance r from the center of the sphere.
  • #1
rkrthegreat
10
0
In the situation shown in the figure a particle having charge q is executing simple harmonic motion of amplitude A and time period T in front of a CONDUCTING EARTHED SPHERE of radius R. The reading of the voltmeter when the charge particle is at its mean position(at a distance r from the sphere's centre) is?

The attempt at a solution
I'm confused about the way to approach the problem. There could be so many ways in which the sphere could get a different potential, (like due to the electric field of the charge q, due to changing magnetic field because of the oscillating charge) that I am unable to get a pinpoint solution. Do help me with this one. Look at the attached diagram for clear details.
 

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  • #2
Is there any other information, such as the resistance of the resistor or how fast the particle moves?
 
  • #3
Oh, I'm sorry, forgot to put that in the diagram. Take the resistance as R0. As for how fast the charge moves, the velocity at any position can be calculated, the amplitude of motion and time period being given in the question.
 
  • #4
I mean, how great is T compared to some other characteristic times, such as r/c (c is speed of light), or the time for charge redistribution? You know, if T is very small compared to them, the problem will be greatly simplified.
 
  • #5
Yes, take T as very small
 
  • #6
Oops sorry. To simplify the problem, T should be very large, not very small. Apology :-p I guess you meant T was large, too? :biggrin:

If T is very large, we may turn the problem to a classical electrostatic one. The field formed by q is electrostatic field. Each position of q corresponds to an electrostatic image on the sphere, which comprises of 2 point charges: one at the center (q1), the other at somewhere midway between the center and q (q2). Denote V the potential of the sphere, then q1 is the one responsible for V: V=kq1/R^2, while q2 is for canceling the potential on the sphere by q, so q2 depends on x and q (I don't remember the formulas, but this can found in most books). Besides, the total charge on the sphere is q1+q2, and its time derivative is the current through the resistor, and thus: V = RoI = -Rod(q1+q2)/dt. You will arrive at a differential equation, and the rest is solving for V :smile: As I remember, the differential equation is quite crazy; but if A<<r, again, we can simplify the equation and it's solvable.
 
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  • #7
Thanks mate, I guess that's what was to be assumed, to reduce the question to a case of electrostatics. That was what I too needed to confirm, otherwise there could have been many paths to take on this question.
(I wrote take T as small because I thought you were proceeding to solve it using electromagnetics ;-) )
 

FAQ: A question on electricity (potential difference)

What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential energy between two points in an electric circuit. It is measured in volts (V) and is the driving force that pushes electric charges through a circuit.

How is potential difference measured?

Potential difference is measured using a voltmeter, which is connected in parallel to the component or circuit being measured. The voltmeter measures the difference in electric potential energy between two points and displays the value in volts.

What is the unit of potential difference?

The unit of potential difference is the volt (V). This unit is named after Italian physicist Alessandro Volta, who invented the first battery.

How does potential difference affect electrical current?

Potential difference is directly proportional to electrical current according to Ohm's Law. This means that as potential difference increases, the current flowing through a circuit also increases. Similarly, as potential difference decreases, the current decreases.

What are some real-world examples of potential difference?

Potential difference is present in many everyday devices and systems, such as batteries, power outlets, and electronic devices. It is also essential for the functioning of household appliances, lighting, and transportation systems, such as trains and cars.

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