1.1 a.) Suppose we have a conductor and place some charge density within it. For a surface S inside the conductor enclose the charge density!
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1 1.1 a. uppose we have a conductor and place some charge density within it. # Q = d 3 x x V ( For a surface inside the conductor enclose the charge density E d a = 1 d 3 x $ %( x$ # V This will cause an electric field that will pull free charges in the conductor to move to cancel the total charge Q from the charge density. Once the total charge is cancelled, there is no more field through. However, ecause the conductor is neutral, pulling a total charge -Q to cancel the charge density will cause a charge Q to develop in the layer immediately surrounding. Call this layer, and let it e infinitesimally aove the surface. 1 of 14
2 1.1 contd a. contd At we have E d a = Q $ Q = # # while at E d a This process will continue until we reach the surface of the conductor, at which point there is no more charge availale to move. Thus the total charge Q ends up distriuted on the surface of the conductor. We also notice that as a result of the process E d a = for any surface inside the conductor. Thus, E = everywhere inside the conductor.. # = Q $ uppose we have a hollow conductor and there is a charge distriution situated outside the conductor: of 14
3 1.1 contd. contd From part a, we know that the charges in the conductor will move until E = everywhere in the volume of the conductor (the region etween and. For any aritrary closed surface in the hollow ecause everywhere in the hollow Thus, E d a # = 1 $ d 3 ( = x ( # x % x = V # E d a = = # V $ E d 3 x meaning that no field lines can start or terminate within the volume of the hollow. o, field lines must pass all the way through in continuous paths. Because we have an electrostatic situation E d l = C for any aritrary closed path C. 3 of 14
4 1.1 contd. contd uppose choose C such part of it runs through the conductor and the other part runs through the hollow, parallel to any aritrary field line: E d l = = C E d l + C E d l Because E = everywhere inside the conductor E d l = C Thus, for any continuous contour through the hollow E d l = E L = # E = C 1 C 1 since L#. Therefore, the electric field is zero everywhere inside the hollow. 4 of 14
5 1.1 contd. contd If a charge density is place inside the hollow then y Gausss Law, for any surface outside the conductor E d a = 1 d 3 x $ %( x $ # & o there must e an electric field outside the conductor. c. V At the surface of a conductor, we can place an aritrarily small loop parallel to the surface E d l = = C E d l + C E d l + C 1 E d l + C a E d l We shrink the curves Ca and C such they are aritrarily small compared to C1 and C. Then E d l = = E in ˆtdl + E out ˆtdl C C C 1 C 5 of 14
6 1.1 contd c. contd However, inside the conductor E in =. Thus, E out ˆtdl = C 1 o there is no component of the electric field tangential to the conductors outer surface. This leaves only the component normal to the outer surface Now if we choose a closed surface intersecting the conductors surface. Further, we choose the surface small enough so that the surface charge density $ is aritrarily close to eing uniform. E d a = 1 d 3 x $ % ( x$ # V E d a = E d a + E d a + E d a + E d a A ince the electric field is zero everywhere inside the conductor E d a = B Thus, we have E d a = E d a + E d a + E d a A B 1 1 We can shift and deform the closed surface such side A is aritrarily close and parallel to the conductors surface, sumerging sides 1 and in the conductor. o, E d a = = E d a 1 E d a = E d a = E n A = 1 da $ # = $ A % E n = $ # # A A 6 of 14
7 1.5 Given the potential ( r = q e $%r 4# r & ' ( From the Poisson equation 1+ %r * + = # $ % we find % 1 $ = # r $r r $ ( & ' $r * + r ( =, 3 e,r However, as r, ( r #. This means there is a point charge at r=. If we choose an aritrarily small sphere of radius r around r=, the potential over that surface can e written ( r = q 1 % 4# r & ' 1+ $r Then we can use Gausss Law over the surface to find the enclosing charge density E d a = 1 d 3 r $ ( r # E = # q ( r = V 4$% r ˆr q = 1 d 3 r ( r # $ ( r = q% ( r V Thus, we find the total charge density as r ( = q r ( # $ 3 e #$r 8% q 8- q ( * = q % 1 4# r + $ & ' ( * 7 of 14
8 1.6 Capacitance is defined y a. C = Q The electric field etween the plates is E = Q A ẑ The potential difference is found on a straight path in the z-direction as Q C = = d $ E # d l = Qd % A Thus, we find the capacitance C = A d 8 of 14
9 1.6 contd. Between the spheres, create the spherical surface with radius r, < r < a. The electric field at the surface is E = Q 4 r ˆr The potential difference along a path from the surface of the inner sphere to the interior surface of the outer sphere is given y = $ E # d l = Q 1 4%& $ r dr = Q ( 1 4%& a ' 1 + *, - = Q 4%& r= r=a a Then, y sustitution of the definition of capacitance, we find ( Q C = Q # a 4 a $ C = 4 a # a ( ( ' a a 9 of 14
10 1.6 contd c. The electric field etween the plates is given y Q E = Lr ˆr Again find the potential difference along a straight path from the inner surface to the outer surface = $ E # d l = r= r=a o find the capacitance Q %& L 1 $ r dr = Q C = Q L ln # & $ % a' ( C = L ln a a ( Q %& L ( ln ' ln a = Q %& L ln ( + * a, - 1 of 14
11 1.1 Apply Greens Theorem to two potentials and : ustitute Poissons equation and continuity across surface oundaries to find d 3 x # $ # ' % ( = & &n $ & * % ( &n +, da = # $ # = $ % % n = # # $ & n = $ % & $ d 3 x # + # $ &, ' ( % % * + = # - # -, ' ( % % * + da $ d 3 x # + $ % da = $ d 3 x # + $ %da 1.13 Consider a point charge q etween the two grounded infinite conductive plates. A surface charge $ is induced on the plates as a result of q, and the potential etween the plates is given y %: ( = q ( x ( y ( z # a ( = ( z = d = r z = 11 of 14
12 1.13 contd If we introduce a corresponding situation in which two infinite conducting plates have opposite surface charge densities $ and -$, giving rise to the potential & etween the plates: ( r = = # V ( z = Vz d d then we can use Greens Reciprocation Theorem as follows to find the total charge induced in the plate of the first situation: $ d 3 x # + da % # $ = d 3 x & = $ d 3 x # + $ da %# $ + $ da %& = q Va d + V da = q Va d + VQ # Q = $q a d 1 of 14
13 1A uppose apply the voltage V to plate at z=d and ground the plate at z=. A charge density flows etween the plates in the negative z-direction at nearly the speed of light. J = cẑ = Jẑ Using Poissons equation = # $ % = J c% The potential is symmetric in the x and y directions. o, x y Thus, otain = # #z = J c$ % = J c$ z This leads to the relation of the applied potential and the current density V = J d J = c V c d 1B i In 1-D, the equation for the Greens function is d dx G x, x with oundary conditions ( = G( x, = G x, Away from x=x, we have d ( = dx G x, x ( = # ( x x which quickly leads to solution G( x, x = Ax < ( x > 13 of 14
14 1B contd Match across x=x x+# $ d x d x G x, x = $ d x % x x x# Thus find ii Use Greens Theorem in 1-D % & # G $ ( # G' ( d x % = *G * x $ G *+ ' &, * x ( - d d d x G x, x ( x> x ( x = x x+# x# d d x G x, x Ax A( x = 1 & A = 1 G( x, x = 1 x < ( x > Applying oundary conditions, we otain ( x + & ( x = 1 # G( x, x% d x% $ & $ x% # ( ( x> x = 'G ' x% ( x < x > x = x = 1 = 1 x ( d x% + x ( + ( x ( ( x ( ( 14 of 14
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