peaceandlove said:Homework Statement
A uniform line charge of density λ lies on the x-axis between x=0 and x=L. Its total charge is 11 nC. The electric field at x=2L is (500 N/C)ˆı. Find the electric field at x=3L. Answer in units of N/C.
Homework Equations
The Attempt at a Solution
I have no clue where to even begin...I know that eventually I need to obtain a relationship between E(2L) and E(3L) but besides that I'm totally lost.
peaceandlove said:I can't seem to get the integral right... I got (1/4(pi)e_0)*(qQ/(d(d+L)). Perhaps I calculated it wrong because I'm not sure what q is.
peaceandlove said:Do I do that using the integral I developed? And if so does that mean the one I came up with is correct?
An electric field is a physical quantity that describes the force exerted on a charged particle by other charged particles. It is a vector quantity, meaning it has both magnitude and direction.
The strength of an electric field is calculated by dividing the force experienced by a charged particle by the magnitude of the charge. It is measured in units of newtons per coulomb (N/C).
Electric field and charge density are directly proportional to each other. This means that as the charge density increases, the electric field also increases, and vice versa. This relationship is described by the equation E = ρ/ε, where E is the electric field, ρ is the charge density, and ε is the permittivity of the medium.
The presence of a charged object creates an electric field in its surroundings. This electric field can either attract or repel other charged objects, depending on their polarity. The strength of the electric field decreases as the distance from the charged object increases.
Conductors have a high concentration of free electrons, which allows them to easily distribute any excess charge and create a uniform electric field. On the other hand, insulators have tightly bound electrons and do not allow for easy distribution of charge, resulting in a non-uniform electric field and higher charge density near the charged object.