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

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SUMMARY

The discussion centers on calculating the electric field inside a parallel-plate capacitor with square plates of side L and distance d between them, where a pendulum of length d/2 and mass m hangs from the top plate. The electric field can be determined using the formula for a parallel-plate capacitor, which is E = σ/ε₀, where σ is the surface charge density. The force acting on the charged mass is not given by F = Bqv, as this pertains to magnetic fields, but rather by F = qE, where E is the electric field calculated earlier. The discussion emphasizes the importance of using the correct equations for electric fields and forces in electrostatics.

PREREQUISITES
  • Understanding of parallel-plate capacitors and their electric fields
  • Familiarity with electrostatic equations, particularly E = σ/ε₀
  • Knowledge of force equations in electrostatics, specifically F = qE
  • Basic calculus for deriving electric fields from charge distributions
NEXT STEPS
  • Research the derivation of the electric field for parallel-plate capacitors
  • Study the relationship between electric field and force on charged particles
  • Explore the concept of surface charge density (σ) and its calculation
  • Review electrostatics equations in textbooks or online resources for clarity on their applications
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or related fields seeking to understand electric fields and forces in capacitors.

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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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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.
 
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
 

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