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Empat muatan masing-masing 6,0 mC ditempatkan pada sudut bujur sangkar dengan sisi 1,0 m. Tentukan besar dan arah gaya yang dialami tiap partikel ....

Empat muatan masing-masing 6,0 mC ditempatkan pada sudut bujur sangkar dengan sisi 1,0 m. Tentukan besar dan arah gaya yang dialami tiap partikel ....

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A. Aulia

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah

jawaban yang tepat adalah 620 comma 2 cross times 10 cubed space straight N

Pembahasan

Diketahui : Pembahasan : Ilustrasi dan garis gaya Kita ambil acuan pada q 4 Panjang garis diagonal pada q 4 dan q 2 adalah m Arah vertikal = sumbu y Arah horisontal = sumbu x Resultan gayanya menjadi : Jadi, jawaban yang tepat adalah

Diketahui :

italic q subscript 1 equals italic q subscript 2 equals italic q subscript 3 equals italic q subscript 4 equals 6 comma 0 space mC equals 6 cross times 10 to the power of negative sign 3 end exponent C italic d equals 1 comma 0 space m

Pembahasan :

Ilustrasi dan garis gaya 

Kita ambil acuan pada q4

Panjang garis diagonal pada q4 dan q2 adalah square root of 2 m

table attributes columnalign right center left columnspacing 0px end attributes row cell q subscript 1 end cell equals cell q subscript 2 equals q subscript 3 equals q subscript 4 equals q equals 6 space mC end cell row cell F subscript 4 comma 3 end subscript end cell equals cell fraction numerator k times q times q over denominator d squared end fraction equals fraction numerator k times q squared over denominator d squared end fraction end cell row cell F subscript 4 comma 1 end subscript end cell equals cell fraction numerator k times q times q over denominator d squared end fraction equals fraction numerator k times q squared over denominator d squared end fraction end cell row cell F subscript 4 comma 2 end subscript end cell equals cell fraction numerator k times q times q over denominator left parenthesis d square root of 2 right parenthesis squared end fraction equals fraction numerator k times q squared over denominator 2 d squared end fraction end cell end table

Arah vertikal = sumbu y

Arah horisontal = sumbu x

F subscript 4 comma 3 subscript X end subscript equals F subscript 4 comma 3 end subscript space cos space 45 F subscript 4 comma 3 subscript y end subscript equals F subscript 4 comma 3 end subscript space cos space 45

 F subscript x equals F subscript 4 comma 3 end subscript plus F subscript 4 comma 2 end subscript space cos space theta F subscript x equals fraction numerator k times q squared over denominator d squared end fraction plus fraction numerator k times q squared over denominator 2 d squared end fraction space cos space 45 F subscript x equals fraction numerator k times q squared over denominator d squared end fraction plus fraction numerator k times q squared over denominator 2 d squared end fraction times 1 half square root of 2 F subscript x equals fraction numerator k times q squared over denominator d squared end fraction open parentheses 1 plus 1 fourth square root of 2 close parentheses F subscript y equals F subscript 4 comma 1 end subscript plus F subscript 4 comma 2 end subscript space cos space theta F subscript y equals fraction numerator k times q squared over denominator d squared end fraction plus fraction numerator k times q squared over denominator 2 d squared end fraction space cos space 45 F subscript y equals fraction numerator k times q squared over denominator d squared end fraction plus fraction numerator k times q squared over denominator 2 d squared end fraction times 1 half square root of 2 F subscript y equals fraction numerator k times q squared over denominator d squared end fraction open parentheses 1 plus 1 fourth square root of 2 close parentheses

Resultan gayanya menjadi :

F equals square root of F subscript x squared plus F end root subscript y squared F equals square root of open parentheses fraction numerator k times q squared over denominator d squared end fraction open parentheses 1 plus 1 fourth square root of 2 close parentheses close parentheses squared plus open parentheses fraction numerator k times q squared over denominator d squared end fraction open parentheses 1 plus 1 fourth square root of 2 close parentheses close parentheses squared end root F equals square root of 2 open parentheses fraction numerator k times q squared over denominator d squared end fraction open parentheses 1 plus 1 fourth square root of 2 close parentheses close parentheses squared end root F equals open parentheses fraction numerator k times q squared over denominator d squared end fraction open parentheses 1 plus 1 fourth square root of 2 close parentheses close parentheses square root of 2 F equals fraction numerator k times q squared over denominator d squared end fraction open parentheses fraction numerator 2 square root of 2 plus 1 over denominator 2 end fraction close parentheses F equals fraction numerator 9 times 10 to the power of 9 times left parenthesis 6 times 10 to the power of negative 3 end exponent right parenthesis squared over denominator 1 squared end fraction open parentheses fraction numerator 2 square root of 2 plus 1 over denominator 2 end fraction close parentheses F equals fraction numerator 9 times 10 to the power of 9 times 36 times 10 to the power of negative 6 end exponent over denominator 1 squared end fraction open parentheses fraction numerator 2 square root of 2 plus 1 over denominator 2 end fraction close parentheses F equals 324 cross times 10 cubed open parentheses 1 comma 9 close parentheses F equals 620 comma 2 cross times 10 cubed space straight N

Jadi, jawaban yang tepat adalah 620 comma 2 cross times 10 cubed space straight N

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