Q1. The electric field is given by the expression E=c/x3 where c is a constant. Find the potential at a point P as a function of x. i.e. find V(x)
Q2. If there is a cavity in a conductor and no charge is placed in the cavity, then prove that the electric field at all the points in the cavity is zero.
1by work energy theoram,
work done by all forces= change in KE of the body...
here, we can measure the potential difference between 2 pts if we move the particle quasistatically, from the first point to the other...
so,
Work done by us(external reagent) + Work done by the E field=∂KE
but here as the process is quasistatic, we mean ∂KE→0
or we have
work done by us=-(work done by E field)
by the definition of a conservative force, W=-∂U
or Work done by us=∂U
Work done by us=∂U= ∫(-E)dx {from infinity to x} as we apply an equivalent and opposite directional force to the E force, so as to keep the process of motion quasistatic,
or
∂U=∫-c/x3dx {infinity to x}
=c/2x2
cheers!!
66The 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.
24Writing my soln again.in a bit more refined way......
consider any 2 points on cavity A and B
Now, B
VB-VA=-∫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,
=>VB-VA=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
9in #8 ur statement
Since any 2 points on conductor are equipotential, is right
but then when dealing inside a cavity its not always right
9if we are getting same result why bother
both methods mite be right but go for shorter one
106so 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.
106that 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:
9anyone trying Q2
its a nice standard concept to know and till now no one has posted xact explanation
1Assuming 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}}
9@ ashish uve applied sign convention wrongly
dx means a small displacement along positive x direction
so E . dx = E.dx
9gordo edits in ur post
replace dU by dU/dq ( WD per unit charge ie dV ) where dq is test charge
also V = c/2x2 + C
now V at ∞ = 0 (by definition) so C= 0
hence V = c/2x2
1Q1 V(x) = - ∫ E dx
Q2 ∫ E dA = Q/ε0 .. where Q is the charge contained in the cavity ...... since Q is 0 ... Electric field is 0 at all pts in the cavity
106im saying that ∫U = -∫E.dr
So, if the particle goes in a direction opposite the electric field. Its pot energy should increase.
Here it is doing so.
So its pot energy shud increase
i.e. ΔU>0
hence ΔV>0.
So potential at that point must be posititve wrt infinity
however if u use wat ive ritten in #3 then u get potential at that point to be -ve.
106cel: y pot has to be -ve? it is moving OPPOSITE to E isnt it? then pot shud increase i.e. is is positive wrt ∞.
9ya pot has to be -ve compared to inf so charged body moves frm - ve to 0 ie x to inf which is correct
106but 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
9sorry i was talking rubbish
field is cylindrically symmetric
24consider any 2 points in cavity A and B
Now,
B
VB-VA=-∫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,
=>VB-VA=0
=> The path integral =0
=> Electric field within cavity is equal to zero at all points in cavity
106but 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?
9#3 field is spherically symmetric so dot product is same as product of magnitudes taking displacement along radial direction
66@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.
