Đáp án:

\(\begin{array}{l}
1.\\
a.\\
{F_{21}} = 7,{2.10^{ - 6}}\\
b.\\
{E_N} = 1800\\
2.\\
a.\\
E = 14V\\
r = 2\Omega \\
b.\\
m = 0,74667g\\
c.\\
{R_1} = 8\Omega 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
a.\\
{F_{21}} = k\frac{{|{q_1}{q_2}|}}{{\varepsilon A{B^2}}} = {9.10^9}\frac{{|{{2.10}^{ - 9}}{{.8.10}^{ - 9}}|}}{{2.0,{1^2}}} = 7,{2.10^{ - 6}}\\
b.\\
{E_A} = k\frac{{|{q_1}|}}{{\varepsilon A{N^2}}} = {9.10^9}.\frac{{{{2.10}^{ - 9}}}}{{2.0,{1^2}}} = 900\\
{E_B} = k\frac{{|{q_2}|}}{{\varepsilon B{N^2}}} = {9.10^9}.\frac{{{{8.10}^{ - 9}}}}{{2.0,{2^2}}} = 900\\
{E_N} = {E_A} + {E_B} = 900 + 900 = 1800\\
2.\\
a.\\
E = 4.3,5 = 14V\\
r = 4.0,5 = 2\Omega \\
b.\\
{R_d} = \frac{{{3^2}}}{3} = 3\Omega \\
R = {R_2} + \frac{{{R_1}{R_d}}}{{{R_1} + {R_d}}} = 6 + \frac{{6.3}}{{6 + 3}} = 10\Omega \\
I = \frac{E}{{R + r}} = \frac{{14}}{{10 + 2}} = \frac{7}{6}A = {I_2}\\
m = \frac{{A{I_2}t}}{{96500.n}} = \frac{{64.\frac{7}{6}.1930}}{{96500.2}} = 0,74667g\\
c.\\
R = {R_2} + \frac{{{R_1}{R_d}}}{{{R_1} + {R_d}}} = 6 + \frac{{{R_1}.3}}{{{R_1} + 3}}\\
I = \frac{E}{{R + r}} = \frac{{14}}{{6 + \frac{{{R_1}.3}}{{{R_1} + 3}} + 2}}\\
I = {I_1} + {I_2} = \frac{{{U_d}}}{{{R_1}}} + \frac{{{U_d}}}{{{R_d}}}\\
\frac{{14}}{{6 + \frac{{{R_1}.3}}{{{R_1} + 3}} + 2}} = \frac{3}{{{R_1}}} + \frac{3}{3}\\
{R_1} = 8\Omega 
\end{array}\)