Answer
(a) $F = 500~N$
(b) $a = 3.0\times 10^{29}~m/s^2$
Work Step by Step
(a) If the proton is 2.0 fm from the surface of the nucleus, the proton is 5.0 fm from the center of the nucleus. Let $q_n$ be the charge of the nucleus and let $q_p$ be the charge of the proton. We can find the electric force on the proton.
$F = \frac{k~q_n~q_p}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)[(54)(1.6\times 10^{-19}~C)](1.6\times 10^{-19}~C)}{(5.0\times 10^{-15}~m)^2}$
$F = 500~N$
(b) We can find the proton's acceleration.
$a = \frac{F}{m}$
$a = \frac{500~N}{1.67\times 10^{-27}~kg}$
$a = 3.0\times 10^{29}~m/s^2$
