How to Set the Upper Plate Potential for an Upward Electric Field?

Click For Summary
SUMMARY

The discussion focuses on calculating the potential required for an upper plate in a parallel plate capacitor setup to create an upward electric field of 4000 V/m, given that the lower plate is at -6V and the plates are separated by 4mm. The relevant equation used is E = V/d, leading to the conclusion that the upper plate must be at -22V to achieve the desired electric field strength. The confusion arises from the direction of the electric field, which necessitates a lower potential on the upper plate compared to the lower plate.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with the equation E = V/d
  • Basic knowledge of calculus for understanding potential gradients
  • Concept of uniform electric fields in parallel plate capacitors
NEXT STEPS
  • Study the derivation of electric field equations in parallel plate capacitors
  • Learn about the relationship between electric field direction and potential difference
  • Explore the implications of voltage polarity in electric field applications
  • Investigate the effects of plate separation on electric field strength
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone interested in understanding electric fields in capacitor systems.

ghulamali
Messages
7
Reaction score
0
complete data:
two large horizontal plates are separated by 4mm .the lower plate is at a potential of -6v.what potential should be applied to upper plate to create an electric field strength of 4000 v/m upward in space between the plates
relavant equation:
E=v/d
attempt at question:
4000=v/0.004
v=16v
but the answer of this question is -22v
how it will be solved
 
Physics news on Phys.org
Umm...I'm not sure what your background is, because this is not advanced physics, yet you have posted it in the advanced physics forum. If you know calculus, then just start from the definition:

\mathbf{E} = -\nabla V = -\mathbf{\hat{x}} \frac{dV}{dx}

Where the last part applies only in this 1-dimensional case, and I've arbitrarily chosen my coordinate system so that the direction along a perpendicular line from lower plate to upper plate is defined to be the positive x-direction. Since the field is uniform, we can simplify further:

|\mathbf{E}| = E = -\frac{\Delta V}{\Delta x}

If you don't know calculus, then ignore what I said before and just start with the formula above. This formula should make sense. The numerator, the change in potential energy per unit charge in moving from one point to another represents the negative of the work done by the electric force over that distance. If you then divide that distance out (the denominator), you just get the electric force per unit charge. Call the voltage on the upper plate V. Then:

\Delta V = V - (-6 \ \textrm{V})

\Delta x = d = 0.004 \ \textrm{m}

E = 4000 \ \frac{\textrm{V}}{\textrm{m}}

Solve for V.
 
Last edited:
ghulamali said:
complete data:
two large horizontal plates are separated by 4mm .the lower plate is at a potential of -6v.what potential should be applied to upper plate to create an electric field strength of 4000 v/m upward in space between the plates
relavant equation:
E=v/d
attempt at question:
4000=v/0.004
v=16v
but the answer of this question is -22v
how it will be solved
You left out thAt the desired field is UPWARD, so the upper plate shoud be at a lower voltage.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 33 ·
2
Replies
33
Views
6K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
6K
  • · Replies 7 ·
Replies
7
Views
20K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K