Question

\begin{itemize} \item Compare the magnitudes of forces for a 10 µm silica particle. \item $F_g = g\rho_p \pi r_p^3 / 6 = 1.2 \times 10^{-11} N = F_r$ \item $F_f = \mu_s F_n = 1.2 \times 10^{-11} N$ (on oxidized wafer). \item $F_d = 3 \mu_a \pi r_d \mu_a = 1.8 \times 10^{-9} N$ ($\mu_a = 1$ m/s). \item $F_i = \rho_a \pi r_d^2 \mu_a^2 / 4 = 9.4 \times 10^{-11} N$ ($\mu_a = 1$ m/s). \item $F_v = N_b q^2 / 4 \pi \epsilon_0 \epsilon_r r^2 = 2.3 \times 10^{-8} N$ (for 100 bonds) or $A_m d_{p1} d_{p2} / ((d_{p1} + d_{p2}) 12r^2) = 4.2 \times 10^{-8} N$ (for two 10 µm particles at 1 nm). \item $F_e = Q_p Q_d / 4 \pi \epsilon_0 \epsilon_r r^2 = 2.3 \times 10^{-10} N$ (for 1 mm separation, $10^6$ charges). \item $F_a = 1.01 \times 10^5 \pi r_d^2 / 2 = 1.6 \times 10^{-5} N$. \end{itemize}What will the particle do under these circumstances?

          \begin{itemize}
\item Compare the magnitudes of forces for a 10 µm silica particle.
\item $F_g = g\rho_p \pi r_p^3 / 6 = 1.2 \times 10^{-11} N = F_r$
\item $F_f = \mu_s F_n = 1.2 \times 10^{-11} N$ (on oxidized wafer).
\item $F_d = 3 \mu_a \pi r_d \mu_a = 1.8 \times 10^{-9} N$ ($\mu_a = 1$ m/s).
\item $F_i = \rho_a \pi r_d^2 \mu_a^2 / 4 = 9.4 \times 10^{-11} N$ ($\mu_a = 1$ m/s).
\item $F_v = N_b q^2 / 4 \pi \epsilon_0 \epsilon_r r^2 = 2.3 \times 10^{-8} N$ (for 100 bonds) or $A_m d_{p1} d_{p2} / ((d_{p1} + d_{p2}) 12r^2) = 4.2 \times 10^{-8} N$ (for two 10 µm particles at 1 nm).
\item $F_e = Q_p Q_d / 4 \pi \epsilon_0 \epsilon_r r^2 = 2.3 \times 10^{-10} N$ (for 1 mm separation, $10^6$ charges).
\item $F_a = 1.01 \times 10^5 \pi r_d^2 / 2 = 1.6 \times 10^{-5} N$.
\end{itemize}What will the particle do under these circumstances?

* Compare the magnitudes of forces for a 10 µm silica particle.

* Fg = gπ rp^3 / 6 = 1.2 × 10^-11 N = Fr

* Ff = Fn = 1.2 × 10^-11 N (on oxidized wafer).

* Fd = 3 π rd = 1.8 × 10^-9 N (= 1 m/s).

* Fi = π rd^2 ^2 / 4 = 9.4 × 10^-11 N (= 1 m/s).

* Fv = Nb q^2 / 4 πϵ0 r^2 = 2.3 × 10^-8 N (for 100 bonds) or Am dp1 dp2 / ((dp1 + dp2) 12r^2) = 4.2 × 10^-8 N (for two 10 µm particles at 1 nm).

* Fe = Qp Qd / 4 πϵ0 r^2 = 2.3 × 10^-10 N (for 1 mm separation, 10^6 charges).

* Fa = 1.01 × 10^5 π rd^2 / 2 = 1.6 × 10^-5 N.
What will the particle do under these circumstances?

Added by Isabel M.

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