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Muatan A 20  (positif) diletakkan pada garis hubung, antara muatan B 20  (negatif) dan muatan C 80  (negatif). Muatan A diletakkan 2 cm dari B dan 4 cm dari C. Jarak BC = 6 cm. Tentukan besarnya resultan gaya yang bekerja pada muatan A!

Pertanyaan

Muatan A 20 μC (positif) diletakkan pada garis hubung, antara muatan B 20 μC (negatif) dan muatan C 80 μC (negatif). Muatan A diletakkan 2 cm dari B dan 4 cm dari C. Jarak BC = 6 cm. Tentukan besarnya resultan gaya yang bekerja pada muatan A!

space space

Pembahasan Video:

Pembahasan Soal:

Diketahui :

table attributes columnalign right center left columnspacing 0px end attributes row cell q subscript A end cell equals cell 20 space μC equals 20 cross times 10 to the power of negative 6 end exponent space straight C end cell row cell q subscript B end cell equals cell negative 20 space μC equals negative 20 cross times 10 to the power of negative 6 end exponent space straight C end cell row cell q subscript C end cell equals cell negative 80 space μC equals negative 80 cross times 10 to the power of negative 6 end exponent space straight C end cell row cell r subscript A B end subscript end cell equals cell 2 space cm equals 0 comma 02 space straight m end cell row cell r subscript B C end subscript end cell equals cell 6 space cm equals 0 comma 06 space straight m end cell row cell r subscript A C end subscript end cell equals cell 4 space cm equals 0 comma 04 space straight m end cell end table 

Ditanyakan: F subscript A equals space... ?
Untuk menetukan besarnya resultan yang bekerja di A, maka dapat digunakan persamaan sebagai berikut.

  1. Memproyeksikan gaya-gaya yang bekerja pada A.

    space space
     
  2. Menentukan resultan gaya yang bekerja pada muatan A.

    F subscript A equals F subscript A C end subscript minus F subscript A B end subscript F subscript A equals fraction numerator k q subscript A q subscript C over denominator r subscript A C end subscript to the power of italic 2 end fraction minus fraction numerator k q subscript A q subscript B over denominator r subscript A B end subscript to the power of italic 2 end fraction F subscript A equals fraction numerator 9 cross times 10 to the power of 9 cross times 20 cross times 10 to the power of negative 6 end exponent cross times 80 cross times 10 to the power of negative 6 end exponent over denominator open parentheses 0 comma 04 close parentheses squared end fraction minus fraction numerator 9 cross times 10 to the power of 9 cross times 20 cross times 10 to the power of negative 6 end exponent cross times 20 cross times 10 to the power of negative 6 end exponent over denominator open parentheses 0 comma 02 close parentheses squared end fraction F subscript A equals fraction numerator 9 cross times 20 cross times 80 cross times 10 to the power of negative 3 end exponent over denominator 16 cross times 10 to the power of negative 4 end exponent end fraction minus fraction numerator 9 cross times 20 cross times 20 cross times 10 to the power of negative 3 end exponent over denominator 4 cross times 10 to the power of negative 4 end exponent end fraction F subscript A equals fraction numerator 14400 cross times 10 to the power of negative 3 end exponent over denominator 16 cross times 10 to the power of negative 4 end exponent end fraction minus fraction numerator 3600 cross times 10 to the power of negative 3 end exponent over denominator 4 cross times 10 to the power of negative 4 end exponent end fraction F subscript A equals open parentheses 900 cross times 10 to the power of negative 3 end exponent close parentheses minus open parentheses 900 cross times 10 to the power of negative 3 end exponent close parentheses F subscript A equals 0 space straight N
     

Jadi, resultan gaya yang bekerja pada muatan A adalah 0 N.space space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 30 Agustus 2021

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Muatan A 20  (positif) diletakkan pada garis hubung, antara muatan B 20  (negatif) dan muatan C 80  (negatif). Muatan A diletakkan 2 cm dari B dan 4 cm dari C. Jarak BC = 6 cm. Tentukan besarnya resul...

Pembahasan Soal:

Diketahui :

q subscript A equals 20 space μC equals 20 cross times 10 to the power of negative 6 end exponent space straight C q subscript B equals negative 20 space μC equals negative 20 cross times 10 to the power of negative 6 end exponent space straight C q subscript C equals negative 80 space μC equals negative 80 cross times 10 to the power of negative 6 end exponent space straight C r subscript A B end subscript equals 2 space cm equals 0 comma 02 space straight m r subscript A C end subscript equals 4 space cm equals 0 comma 04 space straight m r subscript B C end subscript equals 6 space cm equals 0 comma 06 space straight m

Ditanyakan: F subscript B equals space... ?

Resultan gaya Coulomb adalah jumlah dari gaya-gaya Coulomb yang bekerja pada suatu titik tertentu. Untuk menentukan resultan gaya Coulomb kita dapat menggunakan persamaan F equals F subscript 1 plus F subscript 2

Untuk menetukan besarnya resultan yang bekerja di B, maka dapat digunakan persamaan sebagai berikut.

1. Memproyeksikan gaya-gaya yang bekerja pada B.

space space

          F equals F subscript 1 plus F subscript 2 F subscript B equals F subscript B A end subscript plus open parentheses negative F subscript B C end subscript close parentheses F subscript B equals F subscript B A end subscript minus F subscript B C end subscript

2. Menentukan resultan gaya yang bekerja pada muatan B

F subscript B equals F subscript B A end subscript minus F subscript B C end subscript F subscript B equals fraction numerator k q subscript B q subscript A over denominator r subscript A B end subscript to the power of italic 2 end fraction minus fraction numerator k q subscript B q subscript C over denominator r subscript B C end subscript to the power of italic 2 end fraction F subscript B equals fraction numerator 9 cross times 10 to the power of 9 cross times 20 cross times 10 to the power of negative 6 end exponent cross times 20 cross times 10 to the power of negative 6 end exponent over denominator open parentheses 0 comma 02 close parentheses squared end fraction minus fraction numerator 9 cross times 10 to the power of 9 cross times 20 cross times 10 to the power of negative 6 end exponent cross times 80 cross times 10 to the power of negative 6 end exponent over denominator open parentheses 0 comma 06 close parentheses squared end fraction F subscript B equals fraction numerator 3600 cross times 10 to the power of negative 3 end exponent over denominator 4 cross times 10 to the power of negative 4 end exponent end fraction minus fraction numerator 14400 cross times 10 to the power of negative 3 end exponent over denominator 36 cross times 10 to the power of negative 4 end exponent end fraction F subscript B equals fraction numerator open parentheses 900 cross times 10 to the power of negative 3 end exponent close parentheses over denominator 10 to the power of negative 4 end exponent end fraction minus fraction numerator open parentheses 400 cross times 10 to the power of negative 3 end exponent close parentheses over denominator 10 to the power of negative 4 end exponent end fraction F subscript B equals fraction numerator 500 cross times 10 to the power of negative 3 end exponent over denominator 10 to the power of negative 4 end exponent end fraction F subscript B equals 5000 space straight N
 

Jadi, resultan gaya yang bekerja pada muatan B adalah 5000 N ke arah kanan.space space

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Roboguru

Pada titik-titik sudut dari sebuah segitiga sama sisi ditempatkan muatan-muatan listrik q1 = 1  (positif) dan q2 = 2  (positif) dan q3 = 4  (negatif). Jika-panjang sisi-sisi segitiga tersebut 30 cm, t...

Pembahasan Soal:

Diketahui :

q subscript italic 1 equals 1 space μC equals 10 to the power of negative 6 end exponent space straight C q subscript italic 2 equals 2 space μC equals 2 cross times 10 to the power of negative 6 end exponent space straight C q subscript italic 3 equals 4 space μC equals 4 cross times 10 to the power of negative 6 end exponent space straight C r equals 30 space cm equals 0 comma 3 space straight m 

Ditanyakan : F subscript 1 equals... ?
Untuk menentukan besarnya gaya yang bekerja pada muatan q1, maka dapat digunakan persamaan sebagai berikut.

  1. Memproyeksikan gaya-gaya yang bekerja pada muatan q1

    space space
     
  2. Menentukan besarnya gaya yang bekerja pada muatan q1

    F subscript italic 1 equals square root of F subscript italic 12 to the power of italic 2 plus F subscript italic 13 to the power of italic 2 plus 2 F subscript italic 12 F subscript italic 13 cos theta end root F subscript italic 1 equals square root of open parentheses fraction numerator k q subscript italic 1 q subscript italic 2 over denominator r to the power of italic 2 end fraction close parentheses squared plus open parentheses fraction numerator k q subscript italic 1 q subscript italic 3 over denominator r to the power of italic 2 end fraction close parentheses squared plus 2 open parentheses fraction numerator k q subscript italic 1 q subscript italic 2 over denominator r to the power of italic 2 end fraction close parentheses open parentheses fraction numerator k q subscript italic 1 q subscript italic 3 over denominator r to the power of italic 2 end fraction close parentheses cos open parentheses 120 close parentheses end root F subscript italic 1 equals square root of open parentheses fraction numerator 9 cross times 10 to the power of 9 cross times 10 to the power of negative 6 end exponent cross times 2 cross times 10 to the power of negative 6 end exponent over denominator open parentheses 0 comma 3 close parentheses squared end fraction close parentheses squared plus open parentheses fraction numerator 9 cross times 10 to the power of 9 cross times 10 to the power of negative 6 end exponent cross times 4 cross times 10 to the power of negative 6 end exponent over denominator open parentheses 0 comma 3 close parentheses squared end fraction close parentheses squared plus 2 cross times open parentheses fraction numerator 9 cross times 10 to the power of 9 cross times 10 to the power of negative 6 end exponent cross times 2 cross times 10 to the power of negative 6 end exponent over denominator open parentheses 0 comma 3 close parentheses squared end fraction close parentheses cross times open parentheses fraction numerator 9 cross times 10 to the power of 9 cross times 10 to the power of negative 6 end exponent cross times 4 cross times 10 to the power of negative 6 end exponent over denominator open parentheses 0 comma 3 close parentheses squared end fraction close parentheses cross times open parentheses negative 1 half close parentheses end root F subscript italic 1 equals square root of open parentheses fraction numerator 9 cross times 2 cross times 10 to the power of negative 3 end exponent over denominator 0 comma 09 end fraction close parentheses squared plus open parentheses fraction numerator 9 cross times 4 cross times 10 to the power of negative 3 end exponent over denominator 0 comma 09 end fraction close parentheses squared plus 2 cross times open parentheses fraction numerator 9 cross times 2 cross times 10 to the power of negative 3 end exponent over denominator 0 comma 09 end fraction close parentheses cross times open parentheses fraction numerator 9 cross times 4 cross times 10 to the power of negative 3 end exponent over denominator 0 comma 09 end fraction close parentheses cross times open parentheses negative 1 half close parentheses end root F subscript italic 1 equals square root of open parentheses fraction numerator 9 cross times 2 cross times 10 to the power of negative 3 end exponent over denominator 9 cross times 10 to the power of negative 2 end exponent end fraction close parentheses squared plus open parentheses fraction numerator 9 cross times 4 cross times 10 to the power of negative 3 end exponent over denominator 9 cross times 10 to the power of negative 2 end exponent end fraction close parentheses squared plus 2 cross times open parentheses fraction numerator 9 cross times 2 cross times 10 to the power of negative 3 end exponent over denominator 9 cross times 10 to the power of negative 2 end exponent end fraction close parentheses cross times open parentheses fraction numerator 9 cross times 4 cross times 10 to the power of negative 3 end exponent over denominator 9 cross times 10 to the power of negative 2 end exponent end fraction close parentheses cross times open parentheses negative 1 half close parentheses end root F subscript italic 1 equals square root of open parentheses 0 comma 2 close parentheses squared plus open parentheses 0 comma 4 close parentheses squared plus open parentheses 2 cross times 0 comma 2 cross times 0 comma 4 cross times open parentheses negative 1 half close parentheses close parentheses end root F subscript italic 1 equals square root of 0 comma 04 plus 0 comma 16 minus 0 comma 08 end root F subscript italic 1 equals square root of 0 comma 12 end root F subscript italic 1 equals square root of 0 comma 04 space cross times space 3 end root F subscript italic 1 equals square root of 0 comma 04 space end root cross times square root of 3 F subscript italic 1 equals 0 comma 2 square root of 3 space straight N

Jadi, besarnya gaya yang bekerja pada muatan q1 adalah bottom enclose bold 0 bold comma bold 2 square root of bold 3 bold space bold N bold. end enclose

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Dua buah bola kecil bermuatan positif masing-masing q digantung pada tali sutera seperti pada gambar. Apabila massa masing-masing muatan 10 gram, tentukan besarnya muatan q! (g = 10 )

Pembahasan Soal:

Diketahui:

r equals 0 comma 3 space straight m theta equals 90 degree m equals 10 space gr equals 0 comma 01 space kg q subscript 1 equals q subscript 2 equals q  Gaya space Coulomb space pada space kedua space muatan colon F subscript c equals k q squared over r squared  Gaya space berat space pada space masing minus masing space muatan colon w equals m g  Persamaan space kesetimbangan space vertikal colon capital sigma F subscript y equals 0 T sin left parenthesis theta over 2 right parenthesis minus w equals 0 T sin left parenthesis theta over 2 right parenthesis minus m g equals 0 space... left parenthesis straight i right parenthesis  Persamaan space kesetimbangan space horizontal colon capital sigma F subscript x equals 0 T cos left parenthesis theta over 2 right parenthesis minus F subscript c equals 0 T cos left parenthesis theta over 2 right parenthesis minus k q squared over r squared equals 0 space... left parenthesis ii right parenthesis  Dari space left parenthesis straight i right parenthesis space dan space left parenthesis ii right parenthesis space diperoleh fraction numerator m g over denominator sin left parenthesis theta over 2 right parenthesis end fraction equals fraction numerator k q squared over r squared over denominator cos left parenthesis theta over 2 right parenthesis end fraction fraction numerator m g over denominator sin left parenthesis 45 degree right parenthesis end fraction equals fraction numerator k q squared over r squared over denominator cos left parenthesis 45 degree right parenthesis end fraction m g equals k q squared over r squared  Diperoleh colon q equals r square root of fraction numerator m g over denominator k end fraction end root q equals 0 comma 3 cross times square root of fraction numerator 0 comma 01 cross times 10 over denominator 9 cross times 10 to the power of 9 end fraction end root bold italic q bold equals bold 1 bold space bold μC

Jadi, jawaban yang tepat adalah 1 μCspace 

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Roboguru

Dua bola kecil A dan B masing-masing bermuatan listrik Q. Kedua bola ditempatkan terpisah dan tidak dapat bergeser. Kemudian, sebuah bola kecil ringan C yang bermuatan -0,5Q diletakkan tepat di tengah...

Pembahasan Soal:

P e r h a t i k a n space g a m b a r space b e r i k u t.

M u a tan space C space space k e m u d i a n space d i g e s e r space m e n d e k a t i space m u a tan space A space m a k a space j a r a k space m u a tan space A space k e space C space l e b i h space d e k a t space d a r i p a d a space j a r a k space m u a tan space B space k e space C.  S e h i n g g a space colon  r subscript B C end subscript greater than r subscript C A end subscript  D i k e t a h u i space colon  F equals fraction numerator k Q subscript 1 Q subscript 2 over denominator r squared end fraction  A t a u space colon  F tilde 1 over r squared  M a k a space colon  F subscript C B end subscript less than F subscript C A end subscript  H a l space i n i space a k a n space m e n y e b a b k a n space m u a tan space C space a k a n space l e b i h space c e n d e r u n g space t e r t a r i k space m e n d e k a t i space m u a tan space A. space M e s k i p u n space b e g i t u space m u a tan space C space t e t a p space d i t a r i k space o l e h space m u a tan space B space n a m u n space t i d a k space s e k u a t space t a r i k a n space o l e h space m u a tan space A. space J a d i comma space p e r n y a t a a n space s a l a h space d a n space a l a s a n space b e n a r.

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Roboguru

Dua bola kecil A dan B masing-masing bermuatan listrik Q. Kedua bola ditempatkan terpisah dan tidak dapat bergeser. Kemudian, sebuah bola kecil ringan C yang bermuatan -0,5 Q diletakkan tepat di tenga...

Pembahasan Soal:

Perhatikan gambar berikut.

Muatan C digeser mendekati muatan B.


S e h i n g g a space colon  r subscript B C end subscript less than r subscript C A end subscript  D i k e t a h u i space colon  F equals k fraction numerator Q subscript 1 Q subscript 2 over denominator r squared end fraction  A t a u ∶  F tilde space 1 over r squared  M a k a space colon  F subscript C B end subscript greater than F subscript C A end subscript  H a l space i n i space a k a n space m e n y e b a b k a n space m u a tan space C space a k a n space l e b i h space c e n d e r u n g space t e r t a r i k space m e n d e k a t i space m u a tan space B.  space J a d i comma space p e r n y a t a a n space s a l a h space d a n space a l a s a n space s a l a h.

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Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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