Question

Figure 23-50 shows a very large non-conducting sheet that has a uniform surface charge density of $\sigma =-2.00\mu C/m^2$, it also shows a particle of charge$Q=6.00\mu C$, 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 $E_{net}$ of the sheet and particle zero? (c) If $d=0.800m$ , at what coordinate on the x-axis is${\mathit{E}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{0}$?

Short answer

1. The positive coordinate on the x-axis, if$d=0.2m,$, where the net electric field of the sheet and particle has zero value is $+0.691M$.
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 for$E_{eat}=0$ at is$+0.691m$ .

Step-by-step solution

The given data

1. Surface charge density,$\sigma =-2.00\mu C/m^2$
2. Charge of the particle,$Q=6.00\mu C$
3. Separation distances, $d=0.2$ and d=0.8m .

 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,$E=\frac{\left|\sigma \right|}{2_{{\pi }_o}{r}^2}$ (1)

The electric field for a non-conducting sheet,$E=\frac{Q}{4\pi {\epsilon }_O{r}^2}$(2)

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:

$$\begin{gathered} \frac{\left|\sigma \right|}{2{\epsilon }_o}=\frac{Q}{4\pi {\epsilon }_o{r}^2} \\ r=+\sqrt{\frac{3}{2\pi }}m \end{gathered}$$

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

($x=\pm 0.691m$) 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 $x-axis$is$+0.691m$ .

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$-0.691m$.

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 point$x=+0.691$.

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