Electric Field Between Two Parallel Plates

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SUMMARY

The electric field between two parallel plates is determined by the surface charge on the plates and remains constant regardless of the distance between them, as long as the charge is fixed. The formula Epsilon = Delta V / r illustrates that while the electric field strength remains unchanged with increased plate separation, the electric potential difference (Delta V) increases. When the voltage is held constant and the distance changes, the charge on the plates must also adjust. Additionally, the electric field direction is defined based on the force experienced by a positive charge, which is opposite to the force experienced by a negative charge.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with the concepts of electric potential difference
  • Knowledge of the relationship between charge, electric field, and force
  • Basic algebra for manipulating formulas
NEXT STEPS
  • Study the relationship between electric field strength and surface charge density
  • Learn about the concept of electric potential and its calculations
  • Explore the implications of changing plate separation on electric fields
  • Investigate the behavior of electric fields in different configurations, such as capacitors
USEFUL FOR

Physics students, electrical engineers, and anyone interested in understanding electrostatics and electric field behavior in parallel plate systems.

DMac
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[SOLVED] Electric Field Between Two Parallel Plates

My textbook says that the magnitude of the electric field at any point between two plates (except near the edges) depends only on the magnitude of the charge on each plate.

For example, if the magnitude of the electric field between two plates was 3.2 x 10^2 N/C, the field magnitude would not differ if the plate separation were to triple.

However, in the next section, it talks about electric potential difference, and it gives a formula:

Epsilon = Delta V / r, which implies that the magnitude of the electric field between two large parallel plates is dependent on the distance between them.

Could someone please explain to me how these two seemingly contradictory statements make sense?Thanks in advance. =D
 
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DMac said:
For example, if the magnitude of the electric field between two plates was 3.2 x 10^2 N/C, the field magnitude would not differ if the plate separation were to triple.
If you kept the surface charge on the plates fixed, then the field would not change as you separated the plates (within the limits of the approximation). But the voltage difference certainly changes. Delta V = E*d applies here.

However, in the next section, it talks about electric potential difference, and it gives a formula:

Epsilon = Delta V / r, which implies that the magnitude of the electric field between two large parallel plates is dependent on the distance between them.
If you keep the voltage fixed but change the distance you end up changing the charge on the plates.
 
Oh, I get it now. =P Thanks so much!
 
New Question

Oh yeah, I also have a question, but this one's actually dealing with numbers.

A negative charge of 2.4 x 10^-6 C experiences an electric force of magnitude 3.2 N, acting to the left. Calculate the magnitude of the electric field at that point.

I used the formula
Epsilon = Electric Force / Charge and i got the numerical answer to be 1.3 x 10^6 N/C. But, the answer also includes the direction, which is supposed to be
. How does that work? I thought it would be to the left.​
 
Electric field at a point P is defined as the force experienced by a unit positive charge at that point. Hence, if a negative charge experiences a force to the left, a positive charge would experience a force to the right. Hence, the direction of the Electric Field is to the right.
 
Ha, I can't believe I never thought of that. Thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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