solution:
E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})
x=\frac{r}{a}
infinite case,
Ei=\frac{r}{\epsilon0}
\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})
we have to find x when,
ei-e\delta =1% ,y=ei=1/100 ei
or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei
\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}
4x^2+1 =10^4
x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50
\therefore \frac{r}{a}\approx 50
Answer:
a = (V2 - V1) / t = (0 - 14.6) / 3 = -4.87 m/s^2
It might be useful to convert 14.6 m/s to mph
14.6 m/s * 39.37 in/m = 575 in /s
575 in/s / 12 in/ft = 47.9 ft/sec
47.9 ft/s / 88 ft/sec * 60 mph = 32.6 mph
D. Gravitational potential
