Question

Question 8 In the figure a thin glass rod forms a semicircle of radius r 1.67 cm. Charge is uniformly distributed along the rod, with q 1.38 pC in the upper half and -q-1.38 pC in the lower half. What is the magnitude of the electric field at P, the center of the semicircle? Number the tolerance is +/-5% Click if you would like to Show Worlv/s Units N/C or V/m ion: Open Show Work V/m-s N/C.m

Answer 1

Let the linear density of charge, $\lambda$ = q/0.5*pi*R

The charge on the element of arc length, dq = $\lambda$ .dS

Electric field, dE at the center due to the charge, dq is given by,

                      dE = k.dq/R² directed radially outward.

Resolving this along x and y axis.

                 dE_y = dE cos $\theta$ =  k.dq cos $\theta$ /R²

                       = k. $\lambda$ .dScos $\theta$ /R²

But dS = R.d $\theta$

Hence                dE_y= k. $\lambda$ .R.d $\theta$ . cos $\theta$ /R²

                                = (k. $\lambda$ /R).cos $\theta$ . d $\theta$

Integrating from $\theta$ = 0 to $\theta$ = pi/2

                        E_y = (k. $\lambda$ /R).[sin $\theta$ ] for $\theta$ = 0 to $\theta$ = pi/2

                               = k. $\lambda$ /R

Using, $\lambda$ = 2q/pi*R

                         E_y+=  2kq/pi*R²

Similarly for the lower quadrant.

But the x components cancel out. The y components add up.

So, Total El field,    E = 2.E₊ = 4k.q/pi*R²

where, k = 1/4pi* $\varepsilon$ o

Evaluating,

                 E = 4* 9*10⁹ *1.38*10^-12/3.14*(1.67*10^-2)²

E = 56.73 N/C