Consider a pendulum of length d/2 calculate the electric field inside capacitor

In summary: This equation is used to calculate the force on a particle that is subject to a static electric field. In this problem, there is only one charge, so this equation can be used to calculate the force on the charged particle.
  • #1
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Homework Statement



consider a parallel-plate capacitor with square plates of side L and distance d (<<L) between them, charged with charges +Q and -Q. The plates of the capacitor are horizontal, with the lowest lying on the x-y plane, and the orientation is such that their sides are parallel to the x and y axis, respectively

consider a simple pendulum of length d/2 and mass m, hanging vertically from the centre of the top plate, that can oscillate in the x-z plane. Calculate the electric field inside the capacitor, and the electrical force acting on the mass if it has charge q (the field produced by the charge q itself can be neglected, and the polarity of the capacitor is such that the force will act in the negative z direction).

Homework Equations


E=-grad V
=[q/(4pi*epsilon0)][1/(r^2)]*(r-hat)

The Attempt at a Solution



should i just use the above equation, and sort of ignore the r-hat at the end? would the answer to what the electric field is be zero?

then would the electrical force be F=Bqv?
but what would i put in for B and v?
 
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  • #2
Your attempt at a solution is almost completely wrong. Your second equation,

E =[q/(4pi*epsilon0)][1/(r^2)]*(r-hat)

is the equation for the electric field created by a single point charge. If you distribute many electric charges on the surface of an object, such as a capacitor, the electric field is different; not surprisingly, the formula for the electric field depends on the shape of the object. It is possible to use your formula, and some calculus, to calculate the electric field created by charges distributed on any shape. Your book should include an explanation of how to do this. Alternatively, the formula for an electric field inside a parallel-plate capacitor is used very often, so you can probably find the formula in your book or online.

Now, this gives you the electric field, but you still need to find the force on the charged particle. The one you quote, F = Bqv, is not the one you need. That is actually the force from a magnetic field on a moving charged particle. There are many equations that start with "F;" you need to find the one that applies to this problem, which means you have to read the text around the equation, not just the equation itself. Again, the formula you need can be found in your book or Wikipedia.
 
  • #3
I found another equation:

F(subscript 1 2)=[1/(4pi*epsilon)]*[[q(subscript1)*q(subscript2)]/[(r-subscript1 2)^2]]*(r-hat-subscript1 2)

but I'm not sure how this can be used since there is only one charge mentioned in the question and the equation involves two charges
 

FAQ: Consider a pendulum of length d/2 calculate the electric field inside capacitor

1. What is a pendulum of length d/2?

A pendulum of length d/2 is a type of simple harmonic motion system in which a mass is suspended from a fixed point and allowed to swing back and forth. The length d/2 refers to the distance between the point of suspension and the center of mass of the pendulum.

2. How is a pendulum related to an electric field inside a capacitor?

In this scenario, the pendulum represents the movement of a charged particle inside a capacitor. As the pendulum swings, the charged particle moves back and forth, creating an oscillating electric field inside the capacitor.

3. What is the significance of calculating the electric field inside a capacitor?

The electric field inside a capacitor is important because it determines the amount of energy stored in the capacitor. It also affects the behavior and performance of electronic devices that use capacitors.

4. How is the electric field inside a capacitor calculated?

The electric field inside a capacitor can be calculated using the equation E = V/d, where E is the electric field, V is the potential difference between the plates of the capacitor, and d is the distance between the plates.

5. What factors can affect the electric field inside a capacitor?

The electric field inside a capacitor can be affected by factors such as the distance between the plates, the voltage applied to the capacitor, and the dielectric material between the plates. The shape and size of the plates can also play a role in determining the electric field.

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