Penulis: Mr. Benz (Teacher)

Jawaban: B

Gaya tolak-menolak antara dua bola bermuatan berbanding terbalik dengan kuadrat jarak kedua bola.

\(\dfrac{\text{F}_1}{\text{F}_2} = \left(\dfrac{r_2}{r_1}\right)^2\)

\(\dfrac{18 \times 10^{-4}}{\text{F}_2} = \left(\dfrac{0,15}{0,05}\right)^2\)

\(\dfrac{18 \times 10^{-4}}{\text{F}_2} = 3^2\)

\(\dfrac{18 \times 10^{-4}}{\text{F}_2} = 9\)

\(\text{F}_2 = \dfrac{18 \times 10^{-4}}{9}\)

\(\text{F}_2 = 2 \times 10^{-4} \text{ N}\)

Jawaban: C

F adalah besar gaya tarik-menarik antara dua buah benda bermuatan saat berada di udara.

\(\text{F} = k \cdot \dfrac{q \cdot q}{r^2}\)

\(\text{F} = k \cdot \dfrac{q^2}{r^2}\)

F’ adalah besar gaya tarik-menarik antara dua buah benda bermuatan saat berada di medium lain yang memiliki konstanta dielektrik \(\varepsilon_r = 2\)

\(\text{F’} = \dfrac{1}{\varepsilon_r } \cdot k \cdot \dfrac{2q \cdot q}{\left(\dfrac{1}{2}r\right)^2}\)

\(\text{F’} = \dfrac{1}{2} \cdot k \cdot \dfrac{2q^2}{\dfrac{1}{4}r^2}\)

\(\text{F’} = \dfrac{1}{\cancel{2}} \cdot k \cdot \dfrac{\cancel{2} \cdot q^2}{\dfrac{1}{4}r^2}\)

\(\text{F’} = 4 \cdot k \cdot \dfrac{ q^2}{r^2}\)

\(\text{F’} = 4 \cdot \text{F}\)

Jawaban: (A)  x = -35 cm   (B)  Q3 = -180 μC

Karena \(Q_1\) dan \(Q_2\) berbeda jenis muatan, maka \(Q_3\) diletakkan di luar daerah antara \(Q_1\) dan \(Q_2\) dekat yang muatannya paling kecil.

\(Q_3\) diletakkan di sebelah kiri \(Q_1\) agar resultan gaya di \(Q_3\) dapat nol

\(\text{F}_{13} = \text{F}_{23}\)

\(\cancel{k} \cdot \dfrac{Q_1 \cdot \cancel{Q_3}}{r^2_{13}} = \cancel{k} \cdot \dfrac{Q_2 \cdot \cancel{Q_3}}{r^2_{23}}\)

\(\dfrac{20}{x^2} = \dfrac{45}{(15 + x)^2}\)

\(\sqrt{\dfrac{20}{x^2}} = \sqrt{\dfrac{45}{(15 + x)^2}}\)

\(\dfrac{2}{x} = \dfrac{3}{15 + x}\)

\(30 + 2x = 3x\)

\(x = 30\)

Letak \(Q_3\) berada di sebelah kiri muatan \(Q_1\) sejauh 30 cm

atau dengan kata lain

Letak \(Q_3\) pada \(X = -35 \text{ cm}\) (letaknya diukur dari titik asal (0, 0))

Agar resultan gaya di \(Q_1\) juga nol, maka jenis muatan \(Q_3\) harus negatif

\(\text{F}_{31} = \text{F}_{21}\)

\(k \cdot \dfrac{Q_3 \cdot Q_1}{r^2_{31}} =  k \cdot \dfrac{Q_2 \cdot Q_1}{r^2_{21}}\)

\(\dfrac{Q_3}{30^2} = \dfrac{45}{15^2}\)

\(Q_3 = \dfrac{45}{225} \times 900\)

\(Q_3 = 45\times 4\)

\(Q_3 = 180 \mu \text{C}\)

Jadi besar muatan ketiga adalah 180 μC

Jawaban: A 

\(r = 3 \text{ dm} = 3 \times 10^{-1} \text{ m}\)

\(\color{blue}\text{F} = k\dfrac{q_1 \cdot q_2}{r^2}\)

\(4 \text{ N} = 9 \times 10^9\dfrac{q_1 \cdot q_2}{(3 \times 10^{-1})^2}\)

\(4 = \cancel{9} \times 10^9\dfrac{q_1 \cdot q_2}{\cancel{9} \times 10^{-2}}\)

\(4 = 1 \times 10^{11} \cdot q_1 \cdot q_2\)

\(4 \times 10^{-11} = q_1 \cdot q_2\)

\(40 \times 10^{-12} = q_1 \cdot q_2\)

\(40 (\:\mu \text{C})^2 = q_1 \cdot q_2\dotso \color{red} (1)\)

\(q_1 + q_2 = 6 \:\mu \text{C}\dotso \color{red} (2)\)

Dari pilihan jawaban yang tersedia, yang paling mungkin menjadi jawaban tepat adalah A karena kedua muatan kalau dikali menghasilkan angka 40 dan kalau dijumlahkan menghasilkan angka 6

Jawaban: A

Agar muatan uji \(+q\) mengalami gaya total nol, maka muatan uji tersebut harus diletakkan di sebelah kiri muatan \(+Q\).

\(F_{13}\) = gaya tolak-menolak antara muatan \(+q\) dengan \(+Q\)

\(F_{23}\) = gaya tarik-menarik antara muatan \(+q\) dengan \(-2Q\)

Muatan uji \(+q\) mengalami gaya total nol, maka \(\color{blue} F_{13} = F_{23}\)

Note: tanda muatan tidak ikut dalam perhitungan gaya Coulomb

\(F_{13} = F_{23}\)

\(k \cdot \dfrac{Q \cdot q}{x^2} = k \cdot \dfrac{2Q \cdot q}{(2 + x)^2}\)

\(\cancel{k} \cdot \dfrac{\cancel{Q} \cdot \cancel{q}}{x^2} = \cancel{k} \cdot \dfrac{2\cancel{Q} \cdot \cancel{q}}{(2 + x)^2}\)

\(\dfrac{1}{x^2} = \dfrac{2}{(2 + x)^2}\)

\(\sqrt{\dfrac{1}{x^2}} = \sqrt{\dfrac{2}{(2 + x)^2}}\)

\(\dfrac{1}{x} = \dfrac{\sqrt{2}}{2 + x}\)

\(2 + x = \sqrt{2}x\)

\(2 = \sqrt{2}x \:-\:x\)

\(2 = (\sqrt{2} \:-\:1)x\)

\(x = \dfrac{2}{\sqrt{2}\:-\:1} \times \color{blue} \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1}\)

\(x = 2 + 2\sqrt{2}\)

\(x = 2 + \sqrt{8}\)

Posisi muatan uji dalam sumbu X diukur dari koordinat titik asal (0,0) adalah di \(x = -1 – (2 + \sqrt{8}) = – (3 + \sqrt{8})\) meter.

Jawaban: D

\(\text{F}_{\text{BC}} = k \cdot \dfrac{Q_B \times Q_C}{(r_{BC})^2}\)

\(\text{F}_{\text{BC}} = 9 \times 10^{9} \cdot \dfrac{10 \times 10^{-6} \times 30 \times 10^{-6}}{(20 \times 10^{-2})^2}\)

\(\text{F}_{\text{BC}} = 9 \times 10^{9} \cdot \dfrac{300 \times 10^{-12}}{400 \times 10^{-4}}\)

\(\text{F}_{\text{BC}} = \dfrac{27}{4} \times 10^{1}\)

\(\text{F}_{\text{BC}} = \dfrac{135}{2}\text{ N}\)

\(\text{F}_{\text{BA}} = k \cdot \dfrac{Q_B \times Q_A}{(r_{BA})^2}\)

\(\text{F}_{\text{BA}} = 9 \times 10^{9} \cdot \dfrac{10 \times 10^{-6} \times 40 \times 10^{-6}}{(20 \times 10^{-2})^2}\)

\(\text{F}_{\text{BA}} = 9 \times 10^{9} \cdot \dfrac{400 \times 10^{-12}}{400 \times 10^{-4}}\)

\(\text{F}_{\text{BA}} = 90 \text{ N}\)

\(\text{F}_{\text{B}} = \sqrt{ (\text{F}_{\text{BC}})^2 +  (\text{F}_{\text{BA}})^2 }\)

\(\text{F}_{\text{B}} = \sqrt{ ( \dfrac{135}{2})^2 + 90^2 }\)

\(\text{F}_{\text{B}} = \sqrt{  \dfrac{18225}{4} + 8100 }\)

\(\text{F}_{\text{B}} = \sqrt{ \dfrac{50625}{4} }\)

\(\text{F}_{\text{B}} = \dfrac{225}{2}\)

\(\text{F}_{\text{B}} = 112,5 \text{ N}\)

Jawaban: D

Segitiga sama sisi memiliki sudut yang besarnya masing-masing \(60^{\circ}\)

\(\text{F}_{\text{13}}= k\dfrac{Q_1 \cdot Q_3}{r^2}\)

\(\text{F}_{\text{13}}= k\dfrac{Q \cdot Q}{(2a)^2}\)

\(\text{F}_{\text{13}}= k\dfrac{Q^2}{4a^2}\)

\(\text{F}_{\text{23}} = \text{F}_{\text{13}} =\text{F} =  k\dfrac{Q^2}{4a^2}\)

\(\text{F}_{R} = \sqrt{(\text{F}_{\text{23}} )^2 +  (\text{F}_{\text{13}})^2  + 2 \cdot  \text{F}_{\text{23}} \cdot \text{F}_{\text{13}} \cdot \cos 60^{\circ}}\)

\(\text{F}_{R} = \sqrt{(\text{F})^2 +  (\text{F})^2  + 2 \cdot  \text{F}\cdot \text{F} \cdot \dfrac{1}{2}}\)

\(\text{F}_{R} = \sqrt{3\cdot \text{F}^2}\)

\(\text{F}_{R} = \text{F}\cdot \sqrt{3}\)

\(\text{F}_{R} = k\dfrac{Q^2}{4a^2}\cdot \sqrt{3}\)

\(\text{F}_{R} = \dfrac{kQ^2}{4a^2} \sqrt{3}\)