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
[tex]F=qE[/tex]
Considering the charge on the bead as a single point charge, the electric field generated by it is
[tex]E=k_e \frac{Q}{r^2} [/tex]
with [tex]k_e = 8.99\cdot 10^9 Nm^2/C^2[/tex], [tex]Q=+25 nC=25 \cdot 10^{-9}C[/tex] is the charge on the bead. We want to calculate the field at [tex]r=4.0 cm=0.04 m[/tex]:
[tex]E=(8.99\cdot 10^9) \frac{25\cdot 10^{-9}}{(0.04)^2}=1.4\cdot 10^5 V/m [/tex]
The proton has a charge of [tex]q=1.6\cdot 10^{-19}C[/tex], therefore the force exerted on it is
[tex]F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N[/tex]
And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, [tex]m=1.67\cdot 10^{-27} kg[/tex], we have
[tex]F=ma[/tex]
[tex]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 [/tex]
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:
[tex]a=-1.35 \cdot 10^{13} m/s^2[/tex]
