#### Answer

(a) The acceleration of the proton is $1.7\times 10^{14}~m/s^2$ away from the glass bead.
(b) The acceleration of the electron is $3.2\times 10^{17}~m/s^2$ toward the glass bead.

#### Work Step by Step

(a) We can find the magnitude of the electric force on the proton:
$F = \frac{k~q_b~q_p}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(20\times 10^{-9}~C)(1.6\times 10^{-19}~C)}{(0.010~m)^2}$
$F = 2.88\times 10^{-13}~N$
We can find the magnitude of the acceleration.
$a = \frac{F}{m}$
$a = \frac{2.88\times 10^{-13}~N}{1.67\times 10^{-27}~kg}$
$a = 1.7\times 10^{14}~m/s^2$
Since both charges are positive, the force on the proton is directed away from the glass bead. The acceleration of the proton is $1.7\times 10^{14}~m/s^2$ away from the glass bead.
(b) We can find the magnitude of the electric force on the electron:
$F = \frac{k~q_b~q_p}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(20\times 10^{-9}~C)(1.6\times 10^{-19}~C)}{(0.010~m)^2}$
$F = 2.88\times 10^{-13}~N$
We can find the magnitude of the acceleration.
$a = \frac{F}{m}$
$a = \frac{2.88\times 10^{-13}~N}{9.1\times 10^{-31}~kg}$
$a = 3.2\times 10^{17}~m/s^2$
Since the glass bead's charge is positive and the electron's charge is negative, the force on the electron is directed toward the glass bead. The acceleration of the electron is $3.2\times 10^{17}~m/s^2$ toward the glass bead.