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  2. The electric field gradient is the rate at which the electric field falls off, and it is strongest on such edges and lines and points. You can use the gym analogy to see why that is. Imagine the mutual disdain of the students for each other as behaving like spooky spiky hair that extends ghost-like for many meters out from each student.

  3. Why is electric field lines away from (+) and toward (-)?

    physics.stackexchange.com/questions/288172

    The electric field lines extend away from a positive charge. They move forward a negative charge. Let's take parallel plates, which make a uniform electric field.If we take the basic learning, which I mentioned, in accounts, it's very easy to understand that this is the direction of a positive charge object if we put the object between the ...

  4. Why can two (or more) electric field lines never cross?

    physics.stackexchange.com/questions/107171

    The electric field at any point is the sum of all the fields due to each individual charge in the system. The field has a magnitude and a direction. The field lines are a representation of the magnitude and direction of the field over an illustrated area. The field lines point in the direction of the field.

  5. Why can't electrostatic field lines form closed loops?

    physics.stackexchange.com/questions/105592

    Mar 29, 2014 at 9:12. Yes, electrostatic field lines don't form closed loops because ∇ ×E = 0 ∇ → × E → = 0, meaning it is a curl-free vector field. This is a property of a conservative vector field, as it can be expressed as the gradient of some function. (In this case, the electric field being E = −∇V E = − ∇ V. – vs_292.

  6. Why do electric field lines curve at the edges of a uniform...

    physics.stackexchange.com/questions/578146/why-do-electric-field-lines-curve...

    1. Each positive charge in the left plate creates an electric field radially outward away from it, and the total field produced by the plate is the vector sum of each of these individual fields (plus those of the negative charges, but let's focus on the positive ones). At points near the middle of the plate, the charges above it and charges ...

  7. Electric field lines can be taken as continuous curves in a...

    physics.stackexchange.com/questions/317989/electric-field-lines-can-be-taken...

    By convention, electric field lines are said to start from a positive charge and end at a negative charge. As you can see in the above figure, the field lines come to an abrupt stop at the surface of the charge. When there isn't any charge, the electric field lines must be continuous. The only place where they can start or end is at a charge.

  8. Continuity of Electric Field Lines - Physics Stack Exchange

    physics.stackexchange.com/questions/301821/continuity-of-electric-field-lines

    0. Gauss's law says (in integral form) that. ∮E ⋅ dS = Q ϵ, ∮ E → ⋅ d S → = Q ϵ, where Q Q is the charge enclosed by the closed surface on the LHS. This tells us that the net electric flux into or out of a volume depends on the net charge within it. If there is no charge, then the flux in equals the flux out.

  9. Electric field lines properties - Physics Stack Exchange

    physics.stackexchange.com/questions/455137/electric-field-lines-properties

    1. Field lines are a graphical representation of a vector field (you can draw as many lines as you like). In a well drawn three dimensional sketch, the line density will be proportional to the field strength in each region. Electric field lines should start and end on charges with the number of lines proportional to the charge.

  10. Other field lines represent other useful information, like electric field lines represents the motion of a free charge if it were placed on that line (if no other forces were to act on it). Or gravitational field lines representing lines of equal gravitational potential.

  11. The key insight is that a moving charge induces a magnetic field. This magnetic field, combined with the present electric field, gives you the full form of the Lorentz force: →F = q(→v × →B) + q→E. Here you immediately see that there is both a velocity v of the particle and an acceleration hiding away in the force.