Chapter 22, Problem 1P

To determine

The direction of the electric field along the dashed $45°$ lineaway from the intersection point of both the lengths for the L-shaped object.

Explanation of Solution

Given:

The charge is uniformly distributed along the length of the L-shaped object having equal lengths on both sides.

Introduction:

Write the expression for the electric field due to continuous line charge distribution is

  $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}{\displaystyle \int \frac{\lambda dl\widehat{R}}{R^2}}$

Here, ${\epsilon }_0$ is the free space permittivity, $\lambda$ is the uniform linear charge density, R is the distance of the point from the charge distribution and $dl$ is any elemental length on the L-shaped object carrying the charge.

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Draw the diagram to show the different components of the electric field due to the L-shaped object.

Take an arbitrary point, situated a distance $R$ on the dashed line due to any elemental length along the horizontal length.

Write the expression of the electric field on that point.

  ${\overrightarrow{E}}_H=\frac{1}{4\pi {\epsilon }_0}{\displaystyle \int \frac{\lambda dl\widehat{R}}{R^2}}$

Here, $E_H$ is the value of the electric field and the direction will be vertically upwards.

Similarly, take an arbitrary point situated a distance ‘ $R$ ’ on the dashed line due to any elemental length along the vertical length.

Write the expression for the electric field on that point.ss

  ${\overrightarrow{E}}_V=\frac{1}{4\pi {\epsilon }_0}{\displaystyle \int \frac{\lambda dl\widehat{R}}{R^2}}$

Here $E_V$ is the value of the electric field and the direction will be horizontally towards the right of the vertical length.Take the resultant of both the fields.

The direction of the net electric field ( $E_{\text{net}}$ ) will be along the dashed line away from the intersection point of the both lengths.

Conclusion:

Thus, the direction of the electric field along the dashed $45°$ line away from the intersection point of both the lengths for the L-shaped object.