Answer
(a) by Coulomb's Law, the small polar rectangle exerts a force on $P$ whose vertical component is given by
${F_{vert}} = \left( {\frac{{k\rho Qd}}{{{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}} \right)r\Delta r\Delta \theta $
(b) because the integral over the disk is the limit of the Riemann sum of all small polar rectangles, hence
$F = k\rho Qd\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^R \frac{r}{{{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}{\rm{d}}r{\rm{d}}\theta $
Evaluate:
$F = 2\pi k\rho Qd\left( {\frac{1}{d} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right)$
Work Step by Step
(a) Referring to Figure 21, let the distance from the small polar rectangle to $P$ be $\ell $. So,
${\ell ^2} = {r^2} + {d^2}$
Since the charge density $\rho$ is uniform, the charge of the small polar rectangle of size $\Delta r \times \Delta \theta $ is
$q = \rho r\Delta r\Delta \theta $
The particle $P$ has charge $Q$ coulombs. Thus, by Coulomb's Law, the small polar rectangle exerts a force on $P$ given by
${F_P} = k\frac{{qQ}}{{{\ell ^2}}}$
Since ${\ell ^2} = {r^2} + {d^2}$,
${F_P} = k\frac{{qQ}}{{{r^2} + {d^2}}}$
To find the vertical force ${F_{vert}}$ of ${F_P}$, we project ${F_P}$ onto the $z$-axis. Thus,
${F_{vert}} = {F_P}\cos \phi = k\frac{{qQ}}{{{r^2} + {d^2}}}\cos \phi $
From Figure 21, we obtain $\cos \phi = \frac{d}{{\sqrt {{r^2} + {d^2}} }}$.
Therefore, ${F_{vert}} = k\frac{{qQd}}{{{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}$.
Since $q = \rho r\Delta r\Delta \theta $, so ${F_{vert}} = \left( {\frac{{k\rho Qd}}{{{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}} \right)r\Delta r\Delta \theta $.
By symmetry, the horizontal forces cancel out. So, the force on $P$ acts in the vertical direction.
(b) To find the total force on $P$, we decompose the circular disk into an $N \times M$ grid of small polar rectangles ${{\cal R}_{ij}}$. Now, each ${{\cal R}_{ij}}$ exerts a force $\left( {\frac{{k\rho Qd}}{{{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}} \right){r_{ij}}\Delta r\Delta \theta $ on $P$. Thus, the total force is the Riemann sum:
$\mathop \sum \limits_{i = 1}^N \mathop \sum \limits_{j = 1}^M \left( {\frac{{k\rho Qd}}{{{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}} \right){r_{ij}}\Delta r\Delta \theta $
As $N,M \to \infty $, we obtain the double integral for the total force on $P$:
$F = k\rho Qd\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^R \frac{r}{{{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}{\rm{d}}r{\rm{d}}\theta $
Evaluate
$F = k\rho Qd\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^R r{\left( {{r^2} + {d^2}} \right)^{ - 3/2}}{\rm{d}}r{\rm{d}}\theta $
$ = k\rho Qd\left( {\mathop \smallint \limits_0^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_0^R r{{\left( {{r^2} + {d^2}} \right)}^{ - 3/2}}{\rm{d}}r} \right)$
$ = 2\pi k\rho Qd\left( {\mathop \smallint \limits_0^R r{{\left( {{r^2} + {d^2}} \right)}^{ - 3/2}}{\rm{d}}r} \right)$
$ = - 2\pi k\rho Qd\left( {{{\left( {{r^2} + {d^2}} \right)}^{ - 1/2}}|_0^R} \right)$
$ = - 2\pi k\rho Qd\left( {{{\left( {{R^2} + {d^2}} \right)}^{ - 1/2}} - {d^{ - 1}}} \right)$
So, $F = 2\pi k\rho Qd\left( {\frac{1}{d} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right)$.
