Electric field between 2 objects and electric force on object?

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SUMMARY

The discussion focuses on calculating the electric field and force between two parallel metal plates separated by 1.00 cm and connected to a 500-volt battery. The electric field (E) is determined using the formula E = V/d, resulting in E = 5.0 x 104 V/m. The force (F) on a +2.0 nC charge placed between the plates is calculated using F = qE, yielding F = 1.0 x 10-4 N. The width of the plates does not significantly influence the electric field in the center, as the edge effects are negligible in this context.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with the concept of parallel plate capacitors
  • Knowledge of basic electrical equations (V = Ed, F = qE)
  • Ability to convert units (e.g., nanocoulombs to coulombs)
NEXT STEPS
  • Study the concept of electric field lines and their behavior near edges of conductors
  • Learn about the effects of plate area on capacitance and electric field uniformity
  • Explore the relationship between electric field strength and potential difference in capacitors
  • Investigate vector representation of electric forces and fields
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding the principles of electric fields and forces in parallel plate capacitor systems.

megabassdude
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Homework Statement



Suppose two metal plates are separated by 1.00 cm. These plates are wide:
at least 0.5 m square. We hook wires from a battery to each plate, so that there is a
potential of 500 volts between them.

Homework Equations



(A) Find the electric field between the plates, in
the middle where the field is quite uniform.

(B) Suppose we introduce a small object
which has +2.0 nC (nanocoulombs) of charge on it, placing it between the plates, near
the middle where you just found the electric field. Determine the electric force on this
object.

The Attempt at a Solution



(A)

I'm guessing I use the formula VAB = Ed.

Step 1

1.00 cm converted to meters = 0.01

Step 2

VAB = Ed.
500=(E)(0.01).

E = 5.0 x 104 V/m.

Does the width of the plate have any influence though?

(B)

I can't figure out how to do part (B)

I'm thinking I use the formula F = qE.

Step 1

+2.0 nanoCoulomb's converted to Coulombs = +2.0 x 10-9 C

Step 2

F = qE.
F =(2.0 x 10-9 C)(5.0 x 104 V/m)
F = 1.0 x 10-4 N

Again does width of the plate have any influence?
 
Last edited:
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megabassdude said:

Homework Statement



Suppose two metal plates are separated by 1.00 cm. These plates are wide:
at least 0.5 m square.
We hook wires from a battery to each plate, so that there is a potential of 500 volts between them.

Homework Equations



(A) Find the electric field between the plates, in the middle where the field is quite uniform.
Again does width of the plate have any influence?
Welcome to PF!

The electric field is uniform far from the edges of the plates, where the electric field lines are uneven and curved. In this problem, the plates are wide so the area near the edges is small compared to the total. The edge effect does not influence the surface charge density far from the edges, consequently, neither the electric field in the middle.

ehild
 

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Looks good to me - though the force and the electric field are both vectors and you have only given magnitudes. For completeness, you should say something about direction.

The setup describes a parallel plate capacitor - the width of the plates does affect the physics but not the bit you need: that is already accounted for in the electric field. The only effect is that the field is not uniform near the edge.

Note: +1nC charge in a 5V/m field experiences a 5nN force :)
 

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