91Ó°ÊÓ

Figure 23-28 showing a section of three long closed cylinders centered on the same axis.

Charge of cylinder A,$q_A=+3q_0$

The transverse speed of the wave is the displacement of the wave in the given period of its oscillation and the period is given by the reverse of the frequency of the wave. Thus, using the given formulae with the given data, the required values can be calculated.

Formulae:

The electric field in a closed surface relation according to Gauss law,

$âˆ\circledR \stackrel{→}{E}.\stackrel{→}{dS}=\frac{q_{in}}{{∈}_0}$ ….. (i)

The line charge density of a body,

$\mathrm{λ}=\frac{q}{L}$….. (ii)

Here, E is the electric field, s is the displacement, $q_{in}$ is the charge, ${\mathrm{Î\mu }}_0$ is the permittivity of free space, $\mathrm{λ}$ is the line charge density, L is the length.

Using equation (i), you can get the electric field between at any points between two cylindrical surfaces as follows:

Electric field axis being parallel to the axis of area vector and surface area of cylinder,

$S=2\mathrm{Ï€}rl$

Therefore,

$\begin{array}{l}âˆ\circledR \stackrel{→}{E}.\stackrel{→}{dS}=\frac{q_{in}}{{∈}_0}\\ E\left(2\mathrm{Ï€}rl\right)=\frac{\mathrm{λ}l}{{∈}_0}(âˆ\mu \text{Fromequation}\left(ii\right))\\ E=\frac{\mathrm{λ}}{2\mathrm{Ï€}{∈}_{0}r}\end{array}$

Now, the electric field at point 1 is given as:

$E_1=\frac{{\mathrm{λ}}_1}{2\mathrm{π}{∈}_{0}r}$

Thus, it has no contribution from charges of cylinder B and C, and the net field cannot be zero.

Hence, this case is impossible.

Now, the electric field at point 2 is given as:

$E_2=\frac{{\mathrm{λ}}_1}{2\mathrm{π}{∈}_{0}r}+\frac{{\mathrm{λ}}_2}{2\mathrm{π}{∈}_{0}r}$

So, for electric field to be zero, the charges of A and B are related as:

$\begin{array}{l}{\mathrm{λ}}_2=−{\mathrm{λ}}_1\\ \raisebox{1ex}{$q_2$}\!\left/ \!\raisebox{-1ex}{$L$}\right.=−\raisebox{1ex}{$q_1$}\!\left/ \!\raisebox{-1ex}{$L$}\right.(âˆ\mu \text{fromequation}\left(ii\right))\\ q_2=−3q_0\end{array}$

Thus, the charge on cylinder is $−3q_0$and the charge on cylinder C cannot be predicted.

Now, the electric field at point 3 is given as:

$E_2=\frac{{\mathrm{λ}}_1}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}+\frac{{\mathrm{λ}}_2}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}+\frac{{\mathrm{λ}}_3}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}$

So, for electric field to be zero, the charges of A, B and C are related as:

$\begin{array}{l}{\mathrm{λ}}_1+{\mathrm{λ}}_2+{\mathrm{λ}}_3=0\\ \raisebox{1ex}{$q_1$}\!\left/ \!\raisebox{-1ex}{$L$}\right.+\raisebox{1ex}{$q_2$}\!\left/ \!\raisebox{-1ex}{$L$}\right.+\raisebox{1ex}{$q_3$}\!\left/ \!\raisebox{-1ex}{$L$}\right.=0(âˆ\mu \text{fromequation}\left(ii\right))\\ q_2+q_3=−\left(q_1\right)\end{array}$

Hence, due to insufficiency of data, this case is impossible to found.