91影视

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.

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:

鈥淓lectric field of a charged circular rod,鈥� we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc, $\stackrel{鈫�}{E}=\frac{位\sin 胃}{4蟺{蔚}_{0}r}{\mathrm{l}}_{-0}^0$$\stackrel{鈫�}{E}=\frac{位\sin 胃}{4蟺{蔚}_{0}r}{\mathrm{l}}_{-0}^0$ (i)

Where r = radius of the circle.

位 = charge per unit length of the rod.

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

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

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:

$$\begin{gathered} {\stackrel{鈫�}{E}}_{net}=2(-\hat{j})\frac{位\sin 胃}{4蟺{蔚}_{0}R}{l}_{-90掳}^{90掳} \\ =\left(\frac{q}{{蔚}_0{\mathrm{蟺}}^2{R}^{\mathit{2}}}\right)\hat{j} \\ \left|{\stackrel{鈫�}{E}}_{net}\right|=\left(\frac{1.50脳{10}^{-8}\mathrm{C}}{{蔚}_0{\mathrm{蟺}}^2{\left(8.50脳{10}^{-2}\mathrm{m}\right)}^2}\right) \\ =23.8\mathrm{N}/\mathrm{C} \end{gathered}$$

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

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