enow.com Web Search

Search results

  1. Results from the WOW.Com Content Network
  2. electrostatics - Uniform Electric Field - Physics Stack Exchange

    physics.stackexchange.com/questions/173631/uniform-electric-field

    I was going through the first chapter in Resnick Halliday related to electromagnetism and then I came across something called a uniform electric field. It was written that "If E is uniform (that is, constant in magnitude and direction),then the acceleration of the particle is constant" So my question is: 1)How can E be constant in magnitude?

  3. Force acting on a dipole placed in a non-uniform electric field

    physics.stackexchange.com/questions/192120

    Let us consider a non-uniform electric field $\vec{E}$ increasing in the direction left to right. Case 1: Parallel--As seen above, the dipole-with dipole moment $\vec{p}$-is aligned parallel to the direction of increasing non-uniform electric field, $\vec{E}$.

  4. How can a uniform electric field exist? - Physics Stack Exchange

    physics.stackexchange.com/questions/287933/how-can-a-uniform-electric-field-exist

    $\begingroup$ An approximately uniform electric field can be produced between two oppositely charged parallel plates. $\endgroup$ – Farcher Commented Oct 21, 2016 at 15:46

  5. What is the electric field in a parallel plate capacitor?

    physics.stackexchange.com/questions/65191

    When we find the electric field between the plates of a parallel plate capacitor we assume that the electric field from both plates is $${\bf E}=\frac{\sigma}{2\epsilon_0}\hat{n.}$$ The factor of two in the denominator comes from the fact that there is a surface charge density on both sides of the (very thin) plates.

  6. Potential difference in a non-uniform electric field

    physics.stackexchange.com/.../potential-difference-in-a-non-uniform-electric-field

    Let's say I have two points in a non-uniform electric field and those points have the electric potential 10 V and 5 V. If I then would use a voltmeter to determine the voltage between those points would the voltage be turn out to be 10-5= 5 V?

  7. Solved An electron is released from rest in a uniform - Chegg

    www.chegg.com/homework-help/questions-and-answers/electron-released-rest...

    An electron is released from rest in a uniform electric field of magnitude 2.00 x 104 N/C. Calculate the acceleration of the electron. (Ignore gravitation.) ( mass of electron, m e = 9.11 × 10-31 kilograms and e = 1.6 × 10-16 C)

  8. Solved (9\%) Problem 2: A uniform electric field of - Chegg

    www.chegg.com/homework-help/questions-and-answers/9-problem-2-uniform-electric...

    Question: (9\%) Problem 2: A uniform electric field of magnitude 48.1 N/C is parallel to the x axis. A circular loop of radius 22.2 cm is centered at the origin with the normal to the loop pointing 68.4∘ above the x axis. 50% Part (a) Calculate the electric flux in, newton squared meters per coulomb, through the loop.

  9. As time progresses, the electric field spreads out, reaching farther and farther away, "traveling" at the speed of light. This is not travel, however, it is merely delayed effects of the electric field. Considering the charge in the diagram you have given, when its velocity is constant, its electric field contribution is steady, no change.

  10. Why is the electric field between two parallel plates uniform?

    physics.stackexchange.com/questions/435708/why-is-the-electric-field-between...

    The intuitive answer is the following: When you have only one infinite plate the case is the same. If the plate is infinite in lenght, then "there is no spatial scale" in this problem (to an observer the plate looks the same from any height, the charge density does not change), there is no center and there is nothing (no physical features) that can tell you that you are closer or farther from ...

  11. electrostatics - Field between the plates of a parallel plate...

    physics.stackexchange.com/questions/110480

    If we isolate the positive plate without changing its charge distribution, then the electric field due to it alone is E+ = Q/Aε0 (twice that of a conducting plate due to the induced charge). Similarly, the electric field due to the negative plate is E- = Q/Aε0 as well.