Question

In Fig. 22-51, two curved plastic rods, one of charge +qand the other of charge-q, form a circle of radius R=8.50 cm in an x-yplane. The xaxis 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 xaxis) of the electric field Eproduced at P, the center of the circle?

Short 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 isin the -$\hat{j}$direction, or -90^ocounter-clockwise from the +x-axis.

Step-by-step solution

The given data

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

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)

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.

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^ocounter-clockwise from the +x-axis.