Đáp án:

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
a.\\
F = |{q_1}|E = | - {10^{ - 8}}|.4000 = {4.10^{ - 5}}N\\
b.\\
{E_A} = {E_2} = k\frac{{|{q_2}|}}{{A{B^2}}} = {9.10^9}.\frac{{| - {{2.10}^{ - 8}}|}}{{0,{4^2}}} = 1125\\
2.\\
\begin{array}{*{20}{l}}
{a.}\\
{E = {E_1} + {E_2} = 6 + 9 = 15}\\
{r = {r_1} + {r_2} = 2 + 2 = 4\Omega }\\
{b.}\\
{{R_{d1}} = \frac{{U_{dm1}^2}}{{{P_{dm}}}} = \frac{{{{12}^2}}}{{12}} = 12\Omega }\\
{R = \frac{{({R_1} + {R_2}){R_{d1}}}}{{{R_1} + {R_{d1}} + {R_2}}} = \frac{{(18 + 6).12}}{{18 + 6 + 12}} = 8\Omega }\\
{I = \frac{E}{{R + r}} = \frac{{15}}{{8 + 4}} = 1,25A = {I_1} + {I_{d1}}(1)}\\
{{I_1}({R_1} + {R_2}) = {I_d}{R_{d1}}}\\
{{I_1}(6 + 18) = 12{I_2}}\\
{{I_1} = \frac{5}{{12}}A}\\
{m = \frac{{A{I_2}t}}{{96500n}} = \frac{{64.\frac{5}{{12}}.965}}{{96500.2}} = \frac{2}{{15}}g}\\
{c.}\\
{{R_{d2}} = \frac{{{4^2}}}{4} = 4\Omega }\\
{{U_1} + {U_{d2}} = {U_{d1}} \Rightarrow {U_1} = {U_{d1}} - {U_{d2}} = 12 - 4 = 8V}\\
{{I_1} = {I_{d2}} = \frac{4}{4} = 1A}\\
{{R_1} = \frac{{{U_1}}}{{{I_1}}} = \frac{8}{1} = 8\Omega }\\
{}\\
{}\\
{\rm{}}
\end{array}
\end{array}\)