Answer:
Explanation:
Given:
Capacitance, C = 85 pF = 85 × 10⁻¹² F
Resistance, R = 75 MΩ = 75×10⁶Ω
Charge in capacitor at any time 't' is given as:
where,
Q₀ = Maximum charge = CE
E = Initial voltage
t = time
also, Q = CV
V= Final voltage = 90% of E = 0.9E
thus, we have
or
or
taking log both sides, we get
or
or
or
215
I am assuming that those are the width and length. So to find area all you have to do is multiply the two because lw = a. 12 1/2*17 1/5 = 215
25/2*86/5 = 2150/10 = 215
Charge of electron = 1.6×10−¹⁹
(1.6×10−¹⁹)(1×10²) (2e)
= 3.2×10−¹⁷ J
It is like that, except most nails are steel or stainless steel, slowing to rusting process to about 5 years.
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
