Find the force on a charge +q in air at a distance d from an infinite half-plane filled with dielectric of constant K.

Griffiths Chapter 4, example 8.

k = 1/(4 pi epsilon_o)

• Solve this problem by the method of images.
• The boundary plane is the x-z plane, where y=0.
• Above the plane in air, we have region I (y>=0)
• Below the plane in dielectric of relative constant K we have region II (y<=0)

• The boundary between regions I and II is the x-z plane where y=0
• In this plane the parallel component of E must be continuous (we will use the x-component)
• In this plane the perpendicular (y) component of D must be continuous

• Region I is in air, above the dielectric.
• The E and D fields in this region are due to two charges
• the charge q at d above the plane boundary between air and dielectric
• an 'image' charge q' at a distance d' below plane boundary
• both charges are located at x=0.
• q is at y=d and q' is at y=-d'

• Region II is below the x-z plane in the dielectric
• there are no charges in this region so laplace's equation is obeyed
• the E and D fields in region II are due to a single image charge q''
• q'' is located at x=0 and y=+d''.
• (q'' is not in region II, it's above in region I, and laplace's equation is satisfied in region II)
• In region I, at coordinate x,y we have
• Ex = kqx/((d-y)^2+x^2)^(3/2) + kq'x/((d'+y)^2+x^2)^(3/2)
• Ey = -kq(d-y)/((d-y)^2+x^2)^(3/2)+kq'(d'+y)/((d'+y)^2+x^2)^(3/2)
• region I is in air and E is due to 2 charges, so no K is involved
• In region II at coordinate x,y we have
• Ex = kq''x/K/((d''-y)^2+x^2)^(3/2)
• Ey = -kq''(d''+y)/K/((d''-y)^2+x^2)^(3/2)
• In region II we are in dielectric so the effect of q'' is reduced by K
• Ex must be continuous at y=0 for all values of x. (parallel component of E continuous)
• This requires that d=d'=d'' and that q + q' = q''/K
• Dy must be continuous at y=0 for all values of x (normal component of D continuous)
• This requires d=d'=d'' and that -q+q' = -q''
• (Remember that D is due free charges only. It is what E would be with no dielectric.)
• Solving gives q'' = 2Kq/(K+1) and q' = q-q'' = -q(K-1)/(K+1).
• q' is opposite in sign to q, indicating negative bound charge on the surface
• The force on q from the dielectric plane is due to that from q', or kqq'/(2d)^2