Q27P - Fundamentals Of Physics

Question

In Fig. 22-51, two curved plastic rods, one of charge +q and the other of charge-q , form a circle of radius R=8.50 cm in an x-y plane. The x axis passes through both of the connecting points, and the charge is distributed uniformly on both rods. If q=15.0 pC, what are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the electric field E produced at P, the center of the circle?

Step-by-Step Solution

Answer

1. The magnitude of the electric field at the center of the circle is 23.8 N/C.
2. The direction of the electric field at the center of the circle is in the -$\hat{j}$ direction, or -90^o counter-clockwise from the +x-axis.

Step 1: The given data

1. Two curved plastic rods of charges +q and -q form a circle of radius, R=8.50 cm.
2. The charge passing through both the connecting points is q = 15 pC.

Step 2: Understanding the concept of electric field

Using the concept of the electric field of a charged rod, we can get the net electric field of the distribution at the center of the formed circle.

Formulae:

“Electric field of a charged circular rod,” we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc, $\overrightarrow{E}=\frac{\lambda \sin \theta }{4\pi {\epsilon }_{0}r}{\mathrm{l}}_{-0}^0$$\overrightarrow{E}=\frac{\lambda \sin \theta }{4\pi {\epsilon }_{0}r}{\mathrm{l}}_{-0}^0$             (i)

Where r = radius of the circle.

          λ = charge per unit length of the rod.

The length of an arc of a circle, $L=r\theta (\mathrm{\theta } \mathrm{in} \mathrm{radians})$                                       (ii)

The linear density of a distribution, $\lambda =q/L$                                                           (iii)

Step 3: a) Calculations of the electric field at the center of the circle

From symmetry, we see that the net field at P is twice the field caused by the upper semi-circular charge, that is +q (and that it points downward). Substituting the given values and using equations (i), (ii) and (iii), we can get the net electric field as:

$$\begin{gathered} {\overrightarrow{E}}_{net}=2(-\hat{j})\frac{\lambda \sin \theta }{4\pi {\epsilon }_{0}R}{l}_{-90°}^{90°} \\ =\left(\frac{q}{{\epsilon }_0{\mathrm{\pi }}^2{R}^{\mathit{2}}}\right)\hat{j} \\ \left|{\overrightarrow{E}}_{net}\right|=\left(\frac{1.50\times {10}^{-8}\mathrm{C}}{{\epsilon }_0{\mathrm{\pi }}^2{\left(8.50\times {10}^{-2}\mathrm{m}\right)}^2}\right) \\ =23.8 \mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the magnitude value of the net field is 23.8 N/C .

Step 4: b) Calculations of the direction of the electric field at the center of the circle

From the calculations of the part (a), we can get that the net electric field, ${\overrightarrow{E}}_{net}$ points in the -$\hat{j}$ direction, or -90^o counter-clockwise from the +x-axis.