shows two metallic plates A and B placed parallel to each other at a separaton d. A uniform electric field E exists between the plates in the direction from plate B to plate A. Find the potential difference between the plates.

Explanation

Let us take the origin at plate a and x-axis
along the direction from plate A to plate B. We have
`V_B - V_A =-int_(r_A)^(r_B) vecE dvecr = -int_0^d - E dx = Ed.
if we work in Cartesian coordinate system
`E = E_x veci +E_y vec j +E_z veck`
and `dvecr = dx veci +dy vecj +dz veck.`
Thus from
`dV = E_x dx- E_y dy-Ez dz. (i)`
If we change x to x + dx keeping y and z constant,
dy = dz = 0 and from (i),
`|(E_x=-(delV)/(delx)),(E_y=-(delV)/(dely)),(E_z = -(delV)/(delz))|`
The symbols `(del)/(delx), (del)/(dely)` ect. are used to indicate that
while differentiating with respect to one coordinate, the other are kept constant.
if we know the electric field in a region, we can
find the electric using equation and
if we know the electrinc potential in a retgion, we can find the electric field using
Equation may also be written as
`dV = -E dr cos theta`
Where `theta` is the angle between the field `vecE` and the small
displacement `dvecr.` Thus.
`-(dV)/(dr) = E cos theta`
We see that, `-(dV)/(dr)` gives the compnent of the electric
field in the direction of displacement `dvecr.` In we show a small displacement PP' = dr. The electric field
is E making an angle `theta` with PP.' we have
`dV = V(P,) - V(P)`
so that `(V(P) - V(P))/(dr) = E cos theta.`
This gives us a method to get the component fo
the electric field in any given direction if we know the
potential. MOve a small distance dr in the given
direction and see the change dV in the potential. The
component of electric field along that direction is
`(dV)/(dr).`
If we move a distance dr in the direction of the
field, `theta` is zero and ` -(dV)/(dr) = E` is maximum. Thus the
electric field is along the direction in which the potential decrease at the maximum rate.
If a mall displacement `dvecr` perpendicular to the
electric field is considered, `theta = 90^@` and
`dV = -vecE.dvecr = -0.` The potential does not vary in a
direction perpendicular to the electric field.