Q40P Figure 23-50 shows a very large ... [FREE SOLUTION]

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Chapter 23: Q40P (page 682)

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

Short Answer

Expert verified

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

Step by step solution

The given data

Surface charge density, $蟽 = - 2.00 渭 C / m^{2}$
Charge of the particle, $Q = 6.00 渭 C$
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, $E = \frac{|蟽|}{2_{蟺_{o}} r^{2}}$ (1)

The electric field for a non-conducting sheet, $E = \frac{Q}{4 蟺蔚_{O} r^{2}}$ (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:

$\frac{|蟽|}{2 蔚_{o}} = \frac{Q}{4 蟺蔚_{o} r^{2}} r = + \sqrt{\frac{3}{2 蟺}} m$

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

( $x = 卤 0.691 m$ ) is in the 鈥渇orbidden region鈥� between the particle and the sheet. Thus, both values are allowed.

Hence, the value of the positive coordinate on the x-axis $x - a x i s$ is $+ 0.691 m$ .

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 $- 0.691 m$ .

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

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