Q26P - Fundamentals Of Physics

Question

In Fig. 22-50, a thin glass rod forms a semicircle of radius r=5.00 cm  . Charge is uniformly distributed along the rod, with +q=4.50pC   in the upper half and -q= -4.50pC  in the lower half. What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the electric field at P, the center of the semicircle?

Step-by-Step Solution

Answer

Answer:

1. The magnitude of the electric field at P, the center of the semicircle is 20.6 N\C .

       2. The direction of the electric field at P, the center of the semicircle is vertically

             downward in the -J direction, or ${270}^0$counter-clockwise from the +x axis

The given data

• 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.

Understanding the concept of electric field

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 $\lambda$    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,$\lambda =q\backslash L$                                                           (iii)

a) Calculation of the net electric field at the center of the semicircle

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  

b) Calculation of the direction of the net electric field at the center of the semicircle

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.