electrostatics

The fundamental idea one need to keep in mind is the fact the line integral for an electrostatic field over a closed path must be zero: i.e. for a electrostatic field \vec{E}, we have
\oint \vec{E}\boldsymbol{\cdot}\mathrm{d}\vec{\ell}=0
where the circle on the integral signifies integration over a closed path.

This idea leads us to conclude that the electrostatic field in an empty cavity inside a conductor must be zero. For any field line would have to begin and end on the cavity wall, going from a plus charge to a minus charge (as in the figure below)

Letting that field line be part of a closed loop, the rest of which is entirely inside the conductor (where \vec{E}=0) the integral \oint\vec{E}\boldsymbol{\cdot}\mathrm{d}\vec{\ell} is obviously non-zero violating the above fact.

Writing my soln again.in a bit more refined way......

consider any 2 points on cavity A and B
Now, B
V_B-V_A=-∫E.ds
A

where E is electric field existing in cavity &
path A→B is within the cavity

Since any 2 points on conductor are equipotential,
=>V_B-V_A=0
=> The path integral must be equal to zero irrespective of teh path followed b/w points A and B from within the cavity

Since path integral from A→B =0 ,
so Electric field within cavity is equal to zero at all points in cavity

in #8 ur statement

Since any 2 points on conductor are equipotential, is right

but then when dealing inside a cavity its not always right

if we are getting same result why bother

both methods mite be right but go for shorter one

@ cele...Q2 already solved in post#8.....check for errors...

so what is the meaning of dr vector. is it displacement vector. If so in this case shudnt we ACTUALLY write
dr' (disp. vector) = -dr
Hence E.dr' = E.dr'.cosÎ
= E(-dr)(-1)
= Edr
and then integrate

although u will get the same result

I mite be raking up a small issue. but fr me it is important. im sorry if those doubts are silly.

rightwards as i said in #20

that means if we try to derive the potential due to a point charge q at a distance r:
then if we want to derive it from definition, i.e. potential = work done by external agent in a quasistatic process taking reference as ∞ and then integrating the work done as a unit positive charge is brought from infinity to r.

my doubt is this:

Assuming that field is only along the x-axis (because it is given only in terms of x)...

dV = -E.dx

dV = -\frac{c}{x^{3}}dx

\int_{V_{inf}}^{V_{P}}{}dV = -\int_{inf}^{x}{}\frac{c}{x^{3}}dx

V_{P} - V_{inf} = \frac{c}{2x^{2}}

Assuming V_{inf} = 0,

V_{P} = \frac{c}{2x^{2}}

@ ashish uve applied sign convention wrongly

dx means a small displacement along positive x direction

so E . dx = E.dx

i dont have hcv

i donno wat u r saying more clearly ?

if u have HCV please see wroked out problems 16 pg 118

but celestine if #3 is right then we get potential negative. But if u analyse physically, the charged body moves in opposite direction of increasing electric field.So, the potential energy should increase hence potential at any point must be positive

sorry i was talking rubbish

field is cylindrically symmetric

consider any 2 points in cavity A and B
Now,
B
V_B-V_A=-∫E.ds
A
where E is electric field existing in cavity
& path A→B is within the cavity
Since any 2 points on conductor are equipotential,
=>V_B-V_A=0
=> The path integral =0
=> Electric field within cavity is equal to zero at all points in cavity

but celestine : field is NOT spherically symmetric . It is directed only along +ve x-axis. So as we calculate potential by taking the particle from ∞ to that point x. displacement vector is opposite to electric field. hence angle is 180°. So why we write -∫Edx instead of-∫-Edx taking into account the dot product?

#3 field is spherically symmetric so dot product is same as product of magnitudes taking displacement along radial direction

@ankit, your argument is wrong. If Q=0, it simply proves that the flux through a closed surface is zero NOT THE FIELD. If you want to prove from your arguments that the field is zero, you need to take somehow that E outside of the integral.