Question

Figure 23-46 shows a very large nonconducting sheet that has a uniform surface charge density s = -5.10 µC/m^2; it also shows a particle of charge Q = 6.80 µC, at distance d from the sheet. Both are fixed in place. If d = 41.0 cm, at what (a) positive and (b) negative coordinate on the x axis (other than infinity) is the net electric field of the sheet and particle zero? (c) If d = 85.0 cm, at what coordinate is E net = 0? Give your answers in cm.

Answer 1

surface charge density σ = -5.10 µC/m²

                                      = -5.10*10^-6 C/m²

                                IσI = 5.10*10^-6 C/m²

charge Q = 6.8 μC = 6.8*10^-6 C

permitivity ε₀ = 8.85*10^-12 C²/N.m²

for non conducting sheet ,

the electric field E = σ/2ε0
but , the electric field E = kQ/r²

therefore,

               IσI/2ε₀ = kQ/r²

              IσI/2ε₀ = Q/4πε₀r²

              r = [Q/2πIσI]^1/2   ............ (1)

              r = ± 0.4607 m

              r = ± 46.07 cm

..................................................................

if, d = 41 cm ,

i.e, r > d

there is both values are allowed.these are given below

a)

on the positive x- axis,

          r =x = +46.07 cm

b)

on the negative x- axis,

          r = x = -46.07 cm

c)

if d = 85 cm

there is one point x = -46.07 cm is in the forbidden

region between the particle and sheet.

so this is disallowed.

hence, the field cancel only at point x = 46.07 cm