Menghitung Gaya Elektrostatis

\(\text{Q}_1 = 2,4 \:\mu\text{ C} = 2,4 \times 10^{-6} \text{ C}\)

\(\text{Q}_2 = -3,0 \:\mu\text{ C} = -3,0 \times 10^{-6} \text{ C}\)

\(\text{r} = 3 \text{ cm} = 3 \times 10^{-2}\text{ m}\)

\(\text{F} = \text{k}\dfrac{\text{Q}_1\cdot \text{Q}_2}{\text{r}^2}\)

\(\text{F} = 9\times 10^9\cdot \dfrac{2,4 \times 10^{-6}\cdot 3,0 \times 10^{-6}}{(3 \times 10^{-2})^2}\)

\(\text{F} = \cancel{9}\times 10^9\cdot \dfrac{7,2 \times 10^{-12}}{\cancel{9}\times 10^{-4}}\)

\(\text{F} = 7,2 \times 10^{-12 + 9 – (-4)}\)

\(\text{F} = 7,2 \times 10^1 = 72 \text{ N}\)

Karena kedua muatan tidak sejenis, maka yang timbul adalah gaya tarik-menarik

\(\text{F} = \text{k}\dfrac{\text{Q}_1\cdot \text{Q}_2}{\text{r}^2}\)

\(90 = 9\times 10^9\cdot \dfrac{5 \times 10^{-6}\cdot 5 \times 10^{-6}}{\text{r}^2}\)

\(\cancel{9}\times 10^1  = \cancel{9}\times 10^9\cdot \dfrac{25\times 10^{-12}}{\text{r}^2}\:\:\:\:\:\color{blue}\text{kali kedua ruas dengan r}^2\)

\(10\times \text{r}^2 = 25\times 10^{9 – 12}\)

bagi kedua ruas dengan 10

\(\text{r}^2 = 2,5\times 10^{-3}\)

\(\text{r}^2 = 25\times 10^{-4}\)

\(\text{r} = \sqrt{25\times 10^{-4}}\)

\(\text{r} = 5\times 10^{-2}\text{ m}\)

Misal gaya elektrostatisnya sekarang adalah \(\text{F’}\),

\(\dfrac{\text{F}}{\text{F’}} = \dfrac{ \cancel{\text{k}}\:\dfrac{\text{Q}_1\cdot \text{Q}_2}{\text{r}^2}}{ \cancel{\text{k}}\:\dfrac{\text{3Q}_1\cdot \text{3Q}_2}{(\frac{1}{2}\text{r})^2}}\)

\(\dfrac{\text{F}}{\text{F’}} = \dfrac{\dfrac{\text{Q}_1\cdot \text{Q}_2}{\text{r}^2}}{\dfrac{9\text{Q}_1\cdot \text{Q}_2}{\frac{1}{4}\text{r}^2}}\)

\(\dfrac{\text{F}}{\text{F’}} = \dfrac{\cancel{\dfrac{\text{Q}_1\cdot \text{Q}_2}{\text{r}^2}}}{36\cdot\cancel{\dfrac{\cdot\text{Q}_1\cdot \text{Q}_2}{\text{r}^2}}}\)

\(\dfrac{\text{F}}{\text{F’}} = \dfrac{1}{36}\)

Kali silang

\(\text{F’} = 36\text{F}\)

Jadi gaya elektrostatisnya sekarang menjadi 36 kali semua

Karena muatan partikel A dan B tidak sejenis maka posisi partikel C harus diletakkan di luar garis AB dekat dengan muatan yang nilainya kecil (dekat dengan A).

Agar resultan gaya elektrostatis di C sama dengan nol maka \(\text{F}_{\text{CA}} = \text{F}_{\text{CB}}\).

\(\text{F}_{\text{CA}}\) = gaya tolak-menolak antara partikel C dan partikel A

\(\text{F}_{\text{CB}}\) = gaya tarik-menarik antara partikel C dan partikel B

\(\text{F}_{\text{CA}} = \text{F}_{\text{CB}}\)

\(\cancel{\text{k}}\dfrac{\cancel{\text{Q}_c}\cdot \text{Q}_a}{\text{r}_{ca}^2} = \cancel{\text{k}}\dfrac{\cancel{\text{Q}_c}\cdot \text{Q}_b}{\text{r}_{cb}^2}\)

\(\dfrac{9 \mu \text{C}}{x^2} = \dfrac{25 \mu \text{C}}{10^2}\)

Kedua ruas diberi akar

\(\sqrt{\dfrac{9 \mu \text{C}}{x^2}} = \sqrt{\dfrac{25 \mu \text{C}}{10^2}}\)

\(\dfrac{3}{x} = \dfrac{5}{10}\)

Kali silang

\(5x = 30\)

\(x = \dfrac{30}{5} = 6 \text{ cm}\)

Jadi posisi partikel C adalah 6 cm di sebelah kiri partikel A

Karena muatan partikel A dan B sejenis maka posisi partikel C harus diletakkan di antara A dan B.

Agar resultan gaya elektrostatis di C sama dengan nol maka \(\text{F}_{\text{CA}} = \text{F}_{\text{CB}}\).

\(\text{F}_{\text{CA}}\) = gaya tolak-menolak antara partikel C dan partikel A

\(\text{F}_{\text{CB}}\) = gaya tolak-menolak antara partikel C dan partikel B

\(\text{F}_{\text{CA}} = \text{F}_{\text{CB}}\)

\(\cancel{\text{k}}\dfrac{\cancel{\text{Q}_c}\cdot \text{Q}_a}{\text{r}_{ca}^2} = \cancel{\text{k}}\dfrac{\cancel{\text{Q}_c}\cdot \text{Q}_b}{\text{r}_{cb}^2}\)

\(\dfrac{36 \mu \text{C}}{x^2} = \dfrac{49 \mu \text{C}}{(12-x)^2}\)

Kedua ruas diberi akar

\(\sqrt{\dfrac{36 \mu \text{C}}{x^2}} = \sqrt{\dfrac{49 \mu \text{C}}{(12-x)^2}}\)

\(\dfrac{6}{x} = \dfrac{7}{(12-x)}\)

Kali silang

\(7x = 6(12-x)\)

\(7x = 72 – 6x\)

\(7x + 6x = 72\)

\(13x = 72\)

\(x = \dfrac{72}{13} = 5,54 \text{ cm}\)

Jadi posisi partikel C adalah 5,54 cm di sebelah kanan partikel A

\(\text{F}_{\text{BA}}\) = gaya tarik-menarik antara partikel B dan partikel A

\(\text{F}_{\text{BA}} = \text{k}\dfrac{\text{Q}_B\cdot \text{Q}_A}{\text{r}_{\text{BA}}^2}\)

\(\text{F}_{\text{BA}}= 9\times 10^9\cdot \dfrac{2\times 10^{-6}\cdot 1\times 10^{-6}}{(3\times 10^{-2})^2}\)

\(\text{F}_{\text{BA}}= \cancel{9}\times 10^9\cdot \dfrac{2\times 10^{-12}}{\cancel{9}\times 10^{-4}}\)

\(\text{F}_{\text{BA}}= 2\times 10^{9 – 12 – (-4)}\)

\(\text{F}_{\text{BA}} = 2\times 10 = 20\text{ N}\)

\(\text{F}_{\text{BC}}\) = gaya tarik-menarik antara partikel B dan partikel C

\(\text{F}_{\text{BC}}  = \text{k}\dfrac{\text{Q}_B\cdot \text{Q}_C}{\text{r}_{\text{BC}}^2}\)

\(\text{F}_{\text{BC}}  = 9\times 10^9\cdot \dfrac{2\times 10^{-6}\cdot 8\times 10^{-6}}{(4\times 10^{-2})^2}\)

\(\text{F}_{\text{BC}} = 9\times 10^9\cdot \dfrac{\cancel{16}\times 10^{-12}}{\cancel{16}\times 10^{-4}}\)

\(\text{F}_{\text{BC}}  = 9\times 10^{9 – 12 – (-4)}\)

\(\text{F}_{\text{BC}} = 9\times 10 = 90\text{ N}\)

Resultan gaya elektrostatis di B adalah \(\text{F}_{\text{B}}\),

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

\(\text{F}_{\text{B}} = \sqrt{20^2 + 90^2}\)

\(\text{F}_{\text{B}} = \sqrt{400 + 8100}\)

\(\text{F}_{\text{B}} = \sqrt{8500}\)

\(\text{F}_{\text{B}} = 10\sqrt{85}\text{ N}\)

Jadi besar resultan gaya elektrostatis di B adalah \(10\sqrt{85}\text{ N}\)