21.79 •• calc positive charge q is distributed uniformly along the x - axis from x = 0 to x = a. a positive…

21.79 •• calc positive charge q is distributed uniformly along the x - axis from x = 0 to x = a. a positive point charge q is located on the positive x - axis at x = a + r, a distance r to the right of the end of q (fig. p21.79). (a) calculate the x - and y - components of the electric field produced by the charge distribution q at points on the positive x - axis where x > a. (b) calculate the force (magnitude and direction) that the charge distribution q exerts on q. (c) show that if r >> a, the magnitude of the force in part (b) is approximately qq/4πε₀r². explain why this result is obtained. try to solve these problems.

21.79 •• calc positive charge q is distributed uniformly along the x - axis from x = 0 to x = a. a positive point charge q is located on the positive x - axis at x = a + r, a distance r to the right of the end of q (fig. p21.79). (a) calculate the x - and y - components of the electric field produced by the charge distribution q at points on the positive x - axis where x > a. (b) calculate the force (magnitude and direction) that the charge distribution q exerts on q. (c) show that if r >> a, the magnitude of the force in part (b) is approximately qq/4πε₀r². explain why this result is obtained. try to solve these problems.

Answer

Explanation:

Step1: Consider a small charge element

Let's consider a small charge element $dQ$ of length $dx$ at position $x$ on the charged - rod. The linear charge density $\lambda=\frac{Q}{a}$, so $dQ = \lambda dx=\frac{Q}{a}dx$.

Step2: Calculate the electric - field due to the small charge element

The electric - field $d\vec{E}$ due to the small charge element $dQ$ at a point $x = a + r$ on the $x$ - axis is given by Coulomb's law: $d\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{dQ}{(a + r - x)^2}\hat{i}$.

Step3: Integrate to find the total electric - field in the $x$ - direction

$E_x=\int_{0}^{a}\frac{1}{4\pi\epsilon_0}\frac{\frac{Q}{a}dx}{(a + r - x)^2}$. Let $u=a + r - x$, then $du=-dx$. When $x = 0$, $u=a + r$; when $x = a$, $u = r$. So $E_x=\frac{Q}{4\pi\epsilon_0 a}\int_{r}^{a + r}\frac{du}{u^2}=\frac{Q}{4\pi\epsilon_0 a}\left[-\frac{1}{u}\right]_{r}^{a + r}=\frac{Q}{4\pi\epsilon_0 ar}\left(1-\frac{r}{a + r}\right)=\frac{Q}{4\pi\epsilon_0 r(r + a)}$. And $E_y = 0$ since the charge distribution is symmetric about the $x$ - axis.

Step4: Calculate the force on the charge $q$

The force $\vec{F}$ on the charge $q$ is given by $\vec{F}=q\vec{E}$. So $F = qE_x=\frac{Qq}{4\pi\epsilon_0 r(r + a)}$, and the direction is along the positive $x$ - axis.

Step5: Approximate the force for $r\gg a$

When $r\gg a$, $r + a\approx r$. Then $F=\frac{Qq}{4\pi\epsilon_0 r^2}$. This result is obtained because when $r\gg a$, the charged rod of length $a$ can be approximated as a point - charge $Q$ located at the origin (the center - of - charge of the rod for a uniform distribution) for the purpose of calculating the electric force on the charge $q$ at a large distance $r$ from the end of the rod.

Answer:

(a) $E_x=\frac{Q}{4\pi\epsilon_0 r(r + a)}$, $E_y = 0$ (b) $F=\frac{Qq}{4\pi\epsilon_0 r(r + a)}$, direction: along the positive $x$ - axis (c) When $r\gg a$, $r + a\approx r$, so $F=\frac{Qq}{4\pi\epsilon_0 r^2}$. The charged rod can be approximated as a point - charge $Q$ located at the origin for large $r$ relative to $a$.