Test Index

Electrostatics Part 1

Section: Physics

Question : 74 of 100

Marks: +1, -0

r

q

\overset{\rightarrow}{E}

O

Solution:

Let us consider a differential element

d l

. charge on this element.

d q = \left( \; \frac{q}{\pi r} \right) d l

= \; \frac{q}{\pi r} (r d \theta) \;\; (

d l = r d \theta)

= \left( \; \frac{q}{\pi} \right) d \theta

Electric field at

O

due to

d q

\; d E = \; \frac{1}{4 \pi E_0} \cdot \; \frac{d q}{r^2}

\; = \; \frac{1}{4 \pi E_0} \cdot \; \frac{q}{\pi r^2} d \theta

The component

d E \cos \theta

will be counter balanced by another element on left portion.

Hence resultant field at

O

is the resultant of the component

d E \sin \; \theta

only.

\; \therefore E = \int d E \sin \; \theta = \int\limits_{0}^{\pi} \; \frac{q}{4 \pi^2 r^2 E_0} \sin \; \theta d \theta

\; = \; \frac{q}{4 \pi^2 r^2 \in_{0}} \left[-\cos \theta\right]_{0}^{\pi}

\; = \; \frac{q}{4 \pi^2 r^2 \in_{0}} (+1+1)

\; = \; \frac{q}{2 \pi^2 r^2 \in_{0}}

The directions of

E

is towards negative

y

-axis.

\therefore \overset{\rightarrow}{E} = - \; \frac{q}{2 \pi^2 r^2 E_0} \hat{j}

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