Question

The electric field produced by, a thin rod of length L that lies along the z-axis and carries uniformly distributed positive charge q, at a point P located on the perpendicular bisector of the rod (the positive y-axis) a distance y from its center is given by equation.


E= 1/(4πε_0 ) q/(y√(y^2+L^2/4))

(a). Does this equation remain valid if the point P is located at negative y-axis? Explain.
(b) Write an equation similar to above equation if the point P is instead located a distance x from the rod on the positive or negative x-axis.
(c) Write an equation in vector component form for the electric field when point P is located a distance d from the rod on the 45º line that bisects the positive x and y-axes.
(d) Write an equation in vector component form that gives the electric field when point P is located at an arbitrary point x, y anywhere in the xy-plane. Check that the components have the correct signs when the point x, y is located in each of the four quadrants.

Answer 1

Answers:

(a) this equation remain valid if the point P is located at negative y-axis if we put y=-y, because the electric field is a vector quantity and became point out in a negative direction of Y axis. That is

$E(-|y|)=-E(|y|)$

(b) $E= \frac {1}{4πε_0}\frac{q} {x\cdot \sqrt{x^2+L^2/4}}$

(c) $\vec E= \frac {1}{4πε_0}\frac{q\cdot (\hat i\cdot cos(45\degree)+\hat j\cdot sin(45\degree))} {d\cdot \sqrt{d^2+L^2/4}}=\\= \frac {\sqrt{2}}{8πε_0}\frac{q\cdot (\hat i+\hat j)} {d\cdot \sqrt{d^2+L^2/4}}$

(d)$\vec E= \frac {1}{4πε_0}\frac{q\vec r} {r^2\cdot \sqrt{r^2+L^2/4}}$ where $\vec r=x\hat i+y\hat j$ .

Will check

(1) $\vec r=y\hat j$ $->$ $\vec E= \frac {1}{4πε_0}\frac{qy\hat j} {y^2\cdot \sqrt{y^2+L^2/4}} =\hat j \cdot \frac {1}{4πε_0}\frac{q} {y\cdot \sqrt{y^2+L^2/4}}$ ;

(2) $\vec r=-y\hat j$ $->$ $\vec E= \frac {1}{4πε_0}\frac{q (-y)\hat j} {y^2\cdot \sqrt{y^2+L^2/4}} =-\hat j \cdot \frac {1}{4πε_0}\frac{q} {y\cdot \sqrt{y^2+L^2/4}}$ ;

(3) $\vec r=x\hat i$ $->$ $\vec E= \frac {1}{4πε_0}\frac{q x\hat i} {x^2\cdot \sqrt{x^2+L^2/4}} =\hat i \cdot \frac {1}{4πε_0}\frac{q} {x\cdot \sqrt{x^2+L^2/4}}$ ;

(4) $\vec r=-x\hat i$ $->$ $\vec E= \frac {1}{4πε_0}\frac{q (-x)\hat i} {x^2\cdot \sqrt{x^2+L^2/4}} =-\hat i \cdot \frac {1}{4πε_0}\frac{q} {x\cdot \sqrt{x^2+L^2/4}}$