Roboguru

Gambar di bawah ini menunjukkan dua bola identik (m = 0,10 g) bermuatan sama menggantung di ujung tali yang sama panjangnya. Pada posisi seperti gambar, kedua bola itu ternyata seimbang. Berapakah muatan bolanya?

Pertanyaan

Gambar di bawah ini menunjukkan dua bola identik (m = 0,10 g) bermuatan sama menggantung di ujung tali yang sama panjangnya. Pada posisi seperti gambar, kedua bola itu ternyata seimbang. Berapakah muatan bolanya?

Pembahasan Soal:

text m end text subscript text 1 end text end subscript text =m end text subscript text 2 end text end subscript text =0,1 g=10 end text to the power of negative 6 end exponent text kg end text text q end text subscript text 1 end text end subscript text =q end text subscript text 2 end text end subscript text =q end text

Mencari besar masing-masing muatan dengan persamaan tegangan tali pada sumbu x dan sumbu y.

text ΣF end text subscript text x end text end subscript text =0 end text text T cos 60-F=0 end text text T  end text fraction numerator text 1 end text over denominator 2 end fraction text =F end text text T =2F end text

text ΣF end text subscript text y end text end subscript text =0 end text text T sin 60-F=0 end text text T sin 60=F end text text 2F·  end text fraction numerator text 1 end text over denominator 2 end fraction square root of 3 text =m·g end text text F·  end text square root of 3 text =10 end text to the power of negative 6 end exponent text ·10 end text text F = end text fraction numerator text 10 end text to the power of negative 5 end exponent over denominator square root of 3 end fraction text N end text

Menggunakan persamaan hukum Coloumb
text F=k end text fraction numerator text q1 end text times text q2 end text over denominator text r end text squared end fraction fraction numerator text 10 end text to the power of negative 5 end exponent over denominator square root of 3 end fraction text =9×10 end text to the power of 9 text q end text squared over text 0,4 end text squared fraction numerator text 10 end text to the power of negative 5 end exponent over denominator square root of 3 text · end text 9 × 10 to the power of 9 end fraction text = end text text q end text squared over text 16×10 end text to the power of negative 2 end exponent fraction numerator text 10 end text to the power of negative 5 end exponent text ·16×10 end text to the power of negative 2 end exponent over denominator square root of 3 text · end text 9 × 10 to the power of 9 end fraction text =q end text squared fraction numerator text 16×10 end text to the power of negative 16 end exponent over denominator square root of 3 text · end text 9 end fraction text =q end text squared square root of fraction numerator text 16×10 end text to the power of negative 16 end exponent over denominator square root of 3 text · end text 9 end fraction end root text =q end text fraction numerator text 4×10 end text to the power of negative 8 end exponent over denominator 1 comma 31 text · end text 3 end fraction text =q end text fraction numerator text 4×10 end text to the power of negative 8 end exponent over denominator 3 comma 93 end fraction text =q end text 1 comma 01 cross times 10 to the power of negative 8 end exponent text C=q end text

Dengan demikian, besar masing-masing muatan adalah 1 comma 01 cross times 10 to the power of negative 8 end exponent text C. end text

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 20 Mei 2021

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