91影视

Using the concept of the electric potential, we can get the potential at the point in the given ranges for the small infinitesimal charges present in the rod. Again, using this value, we can get the x-component of the electric field. This determines the value of the component at the given distance and also gives the direction of the field. Again, in this concept, we can get that the electric field is independent of y terms and the transverse field is zero.

Formulae:

The electric potential at the point Q due to infinitesimal segment,$dV=\frac{1}{4蟺{蔚}_0}\left(\frac{dq}{r}\right)$ (i)

The charge due to linear charge density,$dq=位dz$ (ii)

The electric field at a point due to potential difference, $E=-\frac{鈭�V}{鈭�x}$ (iii)

A small segment is located at the distance of z from the axis.

Consider an infinitesimal segment dz with a charge dq, which is located at distance z from the origin, then using the charge value of equation (ii) in equation (i) as follows:

$dV=\frac{1}{4蟺{蔚}_0}\left(\frac{位dz}{z-x}\right)$

The electric potential at the point Q due to the entire rod is now given by integrating the above value of the potential as follows:

$\begin{array}{rcl}V& =& \underset{0}{\overset{L}{鈭�}}\frac{1}{4蟺{蔚}_0}\frac{位dz}{z-x}\\ & =& \frac{位}{4蟺{蔚}_0}\underset{0}{\overset{L}{鈭�}}\left(\frac{1}{z-x}\right)dz\\ & =& \frac{位}{4蟺{蔚}_0}{\left[In\left(z-x\right)\right]}_0^L\\ & =& \frac{位}{4蟺{蔚}_0}\left[In\left(L=x\right)\mathit{-}In\left(-x\right)\right]\\ & =& \frac{Q}{4蟺{蔚}_{0}L}\left[In\left(\frac{L-x}{-x}\right)\right]\\ & =& \frac{Q}{4蟺{蔚}_{0}L}\left[In\left(\frac{L-x}{x}\right)\right]...............\left(a\right)\\ & =& \frac{Q}{4蟺{蔚}_{0}L}\left[In\left(\frac{d-L}{d}\right)\right]\\ & =& \left[\frac{\left(9脳{10}^{9}N{m}^2/C^2\right)\left(43.6脳{10}^{-15}C\right)}{\left(0.135m\right)}\right]In\left(\frac{d+0.135m}{d}\right)\\ & =& \left(2.90脳{10}^{-3}V\right)In\left(1+\frac{0.135m}{d}\right)\end{array}$

Hence, the electric potential is $\left(2.90脳{10}^{-3}V\right)In\left(1+\frac{0.135m}{d}\right)$. .

Now, the x-component of the electric field can be given using the value of x = 6.20cm in equation (a) as follows:

$\begin{array}{rcl}{E}_x& =& -\frac{鈭�}{鈭�x}\left(-\frac{Q}{4蟺{蔚}_{0}L}\frac{d}{dx}In\left(\frac{x-L}{x}\right)\right)\\ & =& -\frac{1}{4蟺{蔚}_0}\frac{Q}{L}\frac{x}{x-L}In\left(\frac{1}{x}-\frac{x-L}{x^2}\right)\\ & =& -\frac{1}{4蟺{蔚}_0}\frac{Q}{x\left(x-L\right)}\\ & =& -\frac{\left(9脳{10}^{9}N{m}^2/C\right)\left(43.6脳{10}^{-15}C\right)}{x\left(x+0.135m\right)}\\ & =& -\frac{\left(3.92脳{10}^{-4}N{m}^2/C\right)}{x\left(x+0.135m\right)}...............\left(b\right)\end{array}$

Hence, the magnitude of electric field is $\begin{array}{rcl}& & \frac{\left(3.92脳{10}^{-4}N{m}^2/C\right)}{x\left(x+0.135m\right)}\end{array}$.

From the calculations of part (b), we can see the negative sign of the electric field.

Thus, the value is less than zero.

Hence, the direction of E_x is 180^o relative to the x-axis.

Substituting the given value of in equation (b), we can get the x-component of the electric field as follows:

$\begin{array}{rcl}{E}_x& =& -\frac{\left(3.92脳{10}^{-4}N{m}^2/C\right)}{\left(0.062m\right)\left(0.062m+0.135m\right)}\\ & =& -0.0321N/C\end{array}$

Hence, the electric field at the given distance is -0.0321N/C.

Consider two points of equal infinitesimal distance on either side of P₁ and along a line that is perpendicular to the x-axis.

The transverse component of electric field is given as:

$E_y=\frac{V_1-V_2}{y}$

Since the above two points are situated symmetrically with respect to the rod, the potential are equal at these two points.

$V_1鈮�V_2$

Thus, the transverse component of electric field is 0N/C.