91影视

• Glass rod has a semicircle of radius, r = 5cm
• Charge in the upper half of the rod is +q= +4.50pC , and charge in the lower half of the rod is -q = -4.50pC.

For a charged rod, the amount of charge that is held in the unit length of the rod is known as the linear density of that rod

The field due to a charged distribution on a circular arc,

(i)

where, r is the radius of the circle and $位$ is the charge per unit length of the rod

The length of an arc of a circle,L=$r^0$( 0 in radians) (ii)

The linear density of distribution,$位=q\backslash L$ (iii)

Using equations (ii), (iii) in equation (i), we can see that each charged quarter-circle produces a field of magnitude:

The electric field produced by the positive quarter-circle points at $-{45}^0$, and that of the negative quarter-circle points at$-{135}^0$ from the x-axis. As the magnitude of each field is equal, so the net field will point in the direction making an angle of $-{90}^0$ from the x-axis. Thus, the magnitude of the net field using equation (a) is given as:

Hence, the value of the net electric field is

Using the calculations from part (a), we can say that by the symmetry, the net field points vertically downward in the -j direction, or 270$0^{{}^{}}$ counter-clockwise from the +x axis.