# A charge q = 1 \muC is spread uniformly along the surface of a thin plastic rod of length L =...

## Question:

A charge q = 1 {eq}\mu {/eq}C is spread uniformly along the surface of a thin plastic rod of length L = 1.0m. The rod is placed along the x axist with its center at x=0, y=0. Calculate the MAGNITUDE and DIRECTION of the elctric field created by this charge distribution at point P located at (x = 0.5m, y = 0.5m).

## Electrostatic Field:

Electric field due to a single static point charge particle {eq}Q {/eq} at a distance {eq}r {/eq} is given by the Coulomb's law:

{eq}{\bf E}=\dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r^2}~\hat{\bf r}, {/eq}

where {eq}\hat{\bf r} {/eq} gives the direction of the field and {eq}\epsilon_0 {/eq} is the dielectric constant of the free space.

## Answer and Explanation:

**Given:**

- Total charge on the rod: {eq}~~1~\mu C~~- 1.0\times 10^6 C. {/eq}

- Length of the rod: {eq}~~1.0~ m {/eq}

- Elect

ric field to be determined at the point: {eq}~~P~(x=0.5,y=0.5) {/eq}

The electric field at the point {eq}P {/eq} by an infinitesimal line element {eq}du {/eq} of charged rod, according to Coulomb's law, is:

{eq}dE = \lambda \dfrac{1}{4\pi\epsilon_0}~\dfrac{1}{u^2+(1/4)^2}du, {/eq}

where {eq}0\leq u\leq 1 {/eq} corresponding to {eq}1\geq x\geq 0, {/eq}{eq}\lambda {/eq} is the charge per unit length. The x-and the y- components of the field are:

{eq}dE_x = dE\cos\theta=~~ \lambda \dfrac{1}{4\pi\epsilon_0}~\dfrac{u}{(x^2+(1/2)^2)^{3/2}}~du, \\ dE_y = dE\sin\theta=~~ \lambda \dfrac{1}{4\pi\epsilon_0}~\dfrac{1}{2(u^2+(1/2)^2)^{3/2}}~du. {/eq}

Following relations:

{eq}\cos\theta = \dfrac{u}{\sqrt{u^2+(1/4)}}, {/eq}

and

{eq}\sin\theta =\dfrac{1}{2\sqrt{u^2+(1/4)}}, {/eq}

have been used. Therefore, the net x- and y- components can be written as:

{eq}E_x = ~ \dfrac{1}{4\pi\epsilon_0}~\int_{0}^{1}\dfrac{u}{ (u^2+(1/2)^2)^{3/2}}~du, \\ E_y = ~\dfrac{1}{4\pi\epsilon_0}~\int_{0}^{1} \dfrac{\lambda}{2(u^2+(1/2)^2)^{3/2}}~dx. {/eq}

{eq}E_x = ~ \dfrac{q}{4\pi\epsilon_0}~\times (1-\dfrac{1}{\sqrt{5}})~~=2.04\times 10^5~N/C, {/eq}

and

{eq}E_ y= \dfrac{q}{4\pi\epsilon_0}~{\Large [}\dfrac{2}{\sqrt{u^2+(1/4)^2}}{\Large ]}_{0}^{1}~~= 3.66\times 10^5~N/C. {/eq}

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from UExcel Physics: Study Guide & Test Prep

Chapter 12 / Lesson 4