What Is the Applied Voltage V Between Charged Plates?

In summary: The equation for the electric field between two parallel plates is E = -\frac{dV}{dx}. This is simple to find if you have the information, which you don't since you asked for it. To find the voltage, you use the relation \vec{E}=-\frac{dV}{dx} and substitute in the values for V and d. This gives you the voltage across the capacitor.
  • #1
Inertialforce
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Homework Statement


An object with a charge of +4.0 x 10^-18 C and a mass of 1.1 x 10^-15 kg is held stationary by balanced gravitational and electric forces midway between horizontal charged plates. What is the applied voltage V?


Homework Equations





The Attempt at a Solution


I don't really know how to start this question out. Could someone please tell me what it is that I should start out with or find?
 
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  • #2
With parallel plates you can assume the electric field E is a constant. Look up in your text the E field between parallel plates. This plus the fact that the total of the forces on the particle must add to zero (as vectors) since it is stationary, should make the problem simple.
 
  • #3
jambaugh said:
With parallel plates you can assume the electric field E is a constant. Look up in your text the E field between parallel plates. This plus the fact that the total of the forces on the particle must add to zero (as vectors) since it is stationary, should make the problem simple.

Sorry but I still do not entirely understand, could you show/tell me what equation I should start with or what I should find first? I have now included a diagram that has been given for this question.
 

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  • #4
First you've got to find the strength of the electric field. You know that the particle is in equilibrium and it has a given charge. What is the force experienced by a charged particle in an applied field?

Second, the gravitational force is also acting upon the particle. For the particle to be in equilibrium, net force on it must be zero. This gives you the strength of the electric field.

Using this, you must find the potential difference across the capacitor. As you are not given the dimensions of the capacitor, you would need to use the relation [tex]\vec{E}=-\frac{dV}{dx}[/tex] to find the voltage. As James mentioned, that the electric field is a constant will help you here.

From the expression you obtain for voltage, you will find the potential difference across the capacitor.
 
  • #5
This is a very easy problem...maybe you are just making it harder than it is. Have you a text? It should have the E field of a parallel plate capacitor. If not you can find about a thousand google hits on the topic.

Force due to gravity is [tex]-mg\cdot \hat{k}[/tex] as a vector and where [tex]g = 9.8 \frac{meter}{sec^2}[/tex] and [tex]m[/tex] is the mass of the particle in kilograms.

Force due to an electric field on a charged particle is [tex]q\vec{E}[/tex] where [tex]q[/tex] is the charge of the particle in Coulombs and [tex]\vec{E}[/tex] is the vector electric field in units of Newtons per Coulomb = kg m/(C s^2).

Given, as I said, that the E field is a constant throughout the region between the two parallel plates its value will only be a function of the charge on the plates or equivalently of the voltage applied and the separation.

The total force must be zero and so the magnitudes of the gravitational and electrostatic forces must be equal in magnitude and opposite in direction.
I don't know how I can explain it any better without working the entire problem out.
 

FAQ: What Is the Applied Voltage V Between Charged Plates?

What is electrostatics?

Electrostatics is the study of stationary electric charges and the interactions between them. It is a branch of physics that deals with electric fields, electric potential, and electric energy.

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To solve a hard electrostatics question, you need to first understand the concepts and principles involved. Then, you can use mathematical equations and formulas to calculate electric fields, potential differences, and other relevant quantities. It is also important to carefully analyze the given information and apply the appropriate concepts to the problem.

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Why is electrostatics important?

Electrostatics is important because it helps us understand the behavior of electric charges and their interactions. This knowledge is crucial in many fields, including engineering, physics, and chemistry. It also has practical applications in technology and industry, making our lives more convenient and efficient.

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