Figure 23-50 shows a very large non-conducting sheet that has a uniform surface charge density of σ=-2.00μC/m2, it also shows a particle of chargeQ=6.00μC, at distance dfrom the sheet. Both are fixed in place. If d=0.200m , at what (a) positive and (b) negative coordinate on the xaxis (other than infinity) is the net electric field Enet of the sheet and particle zero? (c) If d=0.800m , at what coordinate on the x-axis isEnet=0?

Short Answer

Expert verified
  1. The positive coordinate on the x-axis, ifd=0.2m,, where the net electric field of the sheet and particle has zero value is +0.691M.
  2. The negative coordinate on the x-axis, if D=0.2m, where the net electric field of the sheet and particle has zero value is -0.691m.
  3. The coordinate on the x-axis forEeat=0 at is+0.691m .

Step by step solution

01

The given data

  1. Surface charge density,σ=-2.00μC/m2
  2. Charge of the particle,Q=6.00μC
  3. Separation distances, d=0.2 and d=0.8m .
02

 Step 2: Understanding the concept of the electric field

Using the concept of the electric field from the electric flux theorem of Gauss's law and the electric field at a point due to charged particles, we can get the required value of the distance of the coordinates on the x-axis.

Formulae:

The electric field at a point due to charged particle,E=σ2πor2 (1)

The electric field for a non-conducting sheet,E=Q4πεOr2(2)

03

a) Calculation of the positive x-coordinate if d = 0.2m

The point where the individual fields cancel cannot be in the region between the sheet and the particle (d<x<0) since the sheet and the particle has opposite-signed charges. The point(s) could be in the region to the right of the particle (x>0)and in the region to the left of the sheet(x<d); this is where the condition can be given using equations (1) and (2) as follows:

σ2εo=Q4πεor2r=+32πm

If d=0.20m(which is less than the magnitude of r found above), then neither of the points

(x=±0.691m) is in the “forbidden region” between the particle and the sheet. Thus, both values are allowed.

Hence, the value of the positive coordinate on the x-axis x-axisis+0.691m .

04

b) Calculation of the negative x-coordinate if d = 0.2m

From the calculations of part (a), the value of the negative coordinate on the x-axis is-0.691m.

05

c) Calculation of the coordinate on the x-axis if d = 0.8m

If, however, d=0.80m (greater than the magnitude of r found above), then oneof the points(x=-0-691m) is in the “forbidden region” between the particle and the sheet and is disallowed. In this part, the fields cancel only at the pointx=+0.691.

Hence, the required value of the x-coordinate is+0.691m .

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