Two large, thin, charged plastic circular plates each of radius R are placed a s

Question

Two large, thin, charged plastic circular plates each of radius R are placed a short distance s apart; s is much smaller than the dimensions of a plate (see figure below). The right-hand plate has a positive charge of +Q evenly distributed over its inner surface (Q is a positive number). The left-hand plate has a negative charge of -2Q evenly distributed over its inner surface. A very thin plastic spherical shell of radius r is placed midway between the plates (and shown in cross section). It has a uniformly distributed positive charge of +q. You can ignore the contributions to the electric field due to the polarization of the thin plastic shell and the thin plastic plates. (Assume the +x direction is to the right and the +y direction is upward Use any variable or symbol stated above along with the following as necessary: eo.) 3r 20 (a) Calculate the x and y components of the electric field at location A, a horizontal distance r/2 to the right of the center of the sphere. (b) Calculate thexand y components of the electric field at location B, a horizontal distance 3r to the left of the center of the sphere.

Explanation / Answer

field at A:

as A is inside the shell, field due to shell=0

fiel due to left side charged plate=charge density/(2*epsilon)

and as the charge is negative, direction is along -ve x axis.

field magnitude=(2*Q/(pi*R^2))/(2*epsilon)

=Q/(pi*epsilon*R^2)

field due to right plate is towards -ve x axis.

field magnitude=charge density/(2*epsilon)

=(Q/(pi*R^2))/(2*epsilon)

=Q/(2*pi*epsilon*R^2)

so net field is aong -ve x axis

magnitude=1.5*Q/(pi*epsilon*R^2)

field at B:

field due to the shell=q/(4*pi*epsilon*9*r^2)

and it is directed along -ve x axis.

field due to left side charged plate=charge density/(2*epsilon)

and as the charge is negative, direction is along -ve x axis.

field magnitude=(2*Q/(pi*R^2))/(2*epsilon)

=Q/(pi*epsilon*R^2)

field due to right plate is towards -ve x axis.

field magnitude=charge density/(2*epsilon)

=(Q/(pi*R^2))/(2*epsilon)

=Q/(2*pi*epsilon*R^2)

so net field is aong -ve x axis

magnitude=(1.5*Q/(pi*epsilon*R^2))+(q/(36*pi*epsilon*r^2))

field at C:

field due to the shell:

field is directed from the center of shell towards point C i.e. along 45 degrees with -ve x axis.

field magnitude=q/(4*pi*epsilon*18*r^2)

so x component =field magnitude*(-cos(45))=-q/(72*sqrt(2)*pi*epsilon*r^2)

y component=field magnitude*sin(45)=q/(72*sqrt(2)*pi*epsilon*r^2)

field due to left side charged plate=charge density/(2*epsilon)

and as the charge is negative, direction is along -ve x axis.

field magnitude=(2*Q/(pi*R^2))/(2*epsilon)

=Q/(pi*epsilon*R^2)

field due to right plate is towards -ve x axis.

field magnitude=charge density/(2*epsilon)

=(Q/(pi*R^2))/(2*epsilon)

=Q/(2*pi*epsilon*R^2)

so total field has x component=-(q/(72*sqrt(2)*pi*epsilon*r^2))-(1.5*Q/(pi*epsilon*R^2))

y component=q/(72*sqrt(2)*pi*epsilon*r^2)

answers are:

part a:

Ex=-1.5*Q/(pi*epsilon*R^2)

Ey=0

part b:

Ex=-(1.5*Q/(pi*epsilon*R^2))-(q/(36*pi*epsilon*r^2))

Ey=0

part c:

Ex=-(q/(72*sqrt(2)*pi*epsilon*r^2))-(1.5*Q/(pi*epsilon*R^2))

Ey=q/(72*sqrt(2)*pi*epsilon*r^2)

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