Question

A straight, non-conducting plastic wire 8.50cm long carries a charge density of +175n c/m distributed uniformly along its length. It is lying on a horizontal table top.

Find the magnitude and direction of the Electric Field that is produced 6.0cm directly above its midpoint.
If the wire is now bent into a circle lying flat on the table, Find the magnitude & direction of the Electric field on a point 6.0cm directly above its center.

Answer 1

Answer on Question #54126, Physics / Other

A straight, non-conducting plastic wire $8.50\mathrm{cm}$ long carries a charge density of $+175~\mathrm{nC / m}$ distributed uniformly along its length. It is lying on a horizontal tabletop.

(a) Find the magnitude and direction of the Electric Field that is produced 6.0cm directly above its midpoint.

(b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the Electric field on a point 6.0cm directly above its center.

Solution:

(a)

Using the notation $\lambda = q / L$ we note that the (infinitesimal) charge on an element $dx$ of the rod contains charge $dq = \lambda dx$ . By symmetry, we conclude that all horizontal field components (due to the $dq$ 's) cancel and we need only "sum" (integrate) the vertical components. Symmetry also allows us to integrate these contributions over only half the rod ( $0 \leq x \leq L / 2$ ) and then simply double the result. In that regard we note that $\sin \theta = R / r$ where $r = \sqrt{x^2 + R^2}$ .

Our element dq produces a differential electric field at point P, which is a distance r from the element. Treating the element as a point charge, we can express the magnitude of dE as

$$d E = \frac {1}{4 \pi \varepsilon_ {0}} \frac {d q}{r ^ {2}}$$

All field is

$$\begin{array}{l} E = 2 \int_ {0} ^ {\frac {L}{2}} \frac {1}{4 \pi \varepsilon_ {0}} \frac {d q}{r ^ {2}} \sin \theta = \frac {2}{4 \pi \varepsilon_ {0}} \int_ {0} ^ {\frac {L}{2}} \frac {\lambda d x}{(x ^ {2} + R ^ {2})} \frac {y}{\sqrt {x ^ {2} + R ^ {2}}} = \frac {\lambda R}{2 \pi \varepsilon_ {0}} \int_ {0} ^ {\frac {L}{2}} \frac {d x}{(x ^ {2} + R ^ {2}) ^ {3 / 2}} \\ = \frac {\lambda R}{2 \pi \varepsilon_ {0}} \frac {x}{R ^ {2} \sqrt {x ^ {2} + R ^ {2}}} \Bigg | _ {0} ^ {L / 2} = \frac {\lambda R}{2 \pi \varepsilon_ {0}} \frac {L / 2}{R ^ {2} \sqrt {(L / 2) ^ {2} + R ^ {2}}} = \frac {\lambda}{2 \pi \varepsilon_ {0}} \frac {L}{R \sqrt {L ^ {2} + 4 R ^ {2}}} \\ \end{array}$$ $$E = \frac {1 7 5 * 1 0 ^ {- 9}}{2 * \pi * 8 . 8 5 * 1 0 ^ {- 1 2}} * \frac {0 . 0 8 5}{0 . 0 6 * \sqrt {0 . 0 8 5 ^ {2} + 4 * 0 . 0 6 ^ {2}}} = 3 0 3 1 8. 2 = 3. 0 3 2 * 1 0 ^ {4} \mathrm {N / C}$$

(b)

The perpendicular components cancel and we need not consider them further. This leaves the parallel components; they all have the same direction, so the net electric field at $P$ is their sum.

The parallel component of shown in figure has magnitude $dE\cos \theta$ . The figure also shows us that

$$\cos \theta = \frac {z}{r} = \frac {z}{\sqrt {z ^ {2} + R ^ {2}}}$$

Let ds be the (arc) length of any differential element of the ring. Since $\lambda$ is the charge per unit (arc) length, the element has a charge of magnitude

$$d q = \lambda d s$$

This differential charge sets up a differential electric field $\mathrm{dE}$ at point $\mathrm{P}$ , which is a distance $\mathrm{r}$ from the element.

$$d E = \frac {1}{4 \pi \varepsilon_ {0}} \frac {\lambda d s}{r ^ {2}} = \frac {1}{4 \pi \varepsilon_ {0}} \frac {\lambda d s}{z ^ {2} + R ^ {2}}$$ $$d E \cos \theta = \frac {z \lambda}{4 \pi \varepsilon_ {0} (z ^ {2} + R ^ {2}) ^ {3 / 2}} d s$$

To add the parallel components produced by all the elements, we integrate around the circumference of the ring, from $s = 0$ to $s = 2 \pi R$ .

$$E = \int d E \cos \theta = \frac {z \lambda}{4 \pi \varepsilon_ {0} (z ^ {2} + R ^ {2}) ^ {3 / 2}} \int_ {0} ^ {2 \pi R} d s = \frac {z \lambda (2 \pi R)}{4 \pi \varepsilon_ {0} (z ^ {2} + R ^ {2}) ^ {3 / 2}} = \frac {z \lambda L}{4 \pi \varepsilon_ {0} (z ^ {2} + L ^ {2} / 4 \pi^ {2}) ^ {3 / 2}}$$ $$E = \frac {0 . 0 6 * 1 7 5 * 1 0 ^ {- 9} * 0 . 0 8 5}{4 * \pi * 8 . 8 5 * 1 0 ^ {- 1 2} * (0 . 0 6 ^ {2} + 0 . 0 8 5 ^ {2} / 4 \pi^ {2}) ^ {3 / 2}} = 3 4 4 9 0. 4 \approx 3. 4 5 * 1 0 ^ {4} \mathrm {N / C}$$

Answer: $3.032 * 10^{4} \mathrm{~N} / \mathrm{C}$ ; $3.45 * 10^{4} \mathrm{~N} / \mathrm{C}$

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