Question

What is the the acceleration of a proton that is 4.0 cm from the center of the bead? input positive value if the acceleration is directed toward the bead and negative if it is directed away from the bead?

Answer 1

Missing detail in the text:
"A small glass bead has been charged to + 25 nC "

Solution
The force exerted on a charge q by an electric field E is given by
$F=qE$
Considering the charge on the bead as a single point charge, the electric field generated by it is
$E=k_e \frac{Q}{r^2}$
with $k_e = 8.99\cdot 10^9 Nm^2/C^2$, $Q=+25 nC=25 \cdot 10^{-9}C$ is the charge on the bead. We want to calculate the field at $r=4.0 cm=0.04 m$:
$E=(8.99\cdot 10^9) \frac{25\cdot 10^{-9}}{(0.04)^2}=1.4\cdot 10^5 V/m$
The proton has a charge of $q=1.6\cdot 10^{-19}C$, therefore the force exerted on it is
$F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N$

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, $m=1.67\cdot 10^{-27} kg$, we have
$F=ma$
$a= \frac{F}{m}= \frac{2.25\cdot 10^{-14} N}{1.67\cdot 10^{-27} kg}=1.35 \cdot 10^{13} m/s^2$

The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
$a=-1.35 \cdot 10^{13} m/s^2$