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Segitiga ABC mempunyai panjang sisi 5 cm. Jika di titik sudut A , B dan C terdapat muatan masing-masing sebesar 1 μ C ,2 dan 3 ,hitunglah energi potensial pada ketiga titik itu!

Segitiga ABC mempunyai panjang sisi 5 cm. Jika di titik sudut A, B dan C terdapat muatan masing-masing sebesar 1 , 2 μC dan 3 μC, hitunglah energi potensial pada ketiga titik itu!space space

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Jawaban terverifikasi

Jawaban

energi potesial pada titik A , B , C , dan energi potensial sistem berturut-turut adalah 0,9 J, 1,44 J, 1,62 J, dan 1,98 J.

energi potesial pada titik A, B, C, dan energi potensial sistem berturut-turut adalah 0,9 J, 1,44 J, 1,62 J, dan 1,98 J.space space

Pembahasan

Diketahui : Untuk menentukan energi potensial pada ketiga titik itu, maka dapat digunakan persamaan sebagai berikut. Menentukan energi potensial di titik A Menentukan energi potensial di titik B Menentukan energi potensial di titik C Menentukan energi potensial sistem Jadi, energi potesial pada titik A , B , C , dan energi potensial sistem berturut-turut adalah 0,9 J, 1,44 J, 1,62 J, dan 1,98 J.

Diketahui :

q subscript A equals 1 space μC equals 10 to the power of negative 6 end exponent space straight C q subscript B equals 2 space μC equals 2 cross times 10 to the power of negative 6 end exponent space straight C q subscript C equals 3 space μC equals 3 cross times 10 to the power of negative 6 end exponent space straight C r subscript A B end subscript equals 5 space cm equals 0 comma 05 space straight m r subscript B C end subscript equals 5 space cm equals 0 comma 05 space straight m r subscript C A end subscript equals 5 space cm equals 0 comma 05 space straight m


Untuk menentukan energi potensial pada ketiga titik itu, maka dapat digunakan persamaan sebagai berikut.

  1. Menentukan energi potensial di titik A

    E P subscript A equals E P subscript A B end subscript plus E P subscript A C end subscript E P subscript A equals fraction numerator k q subscript A q subscript B over denominator r subscript A B end subscript end fraction plus fraction numerator k q subscript A q subscript C over denominator r subscript A C end subscript end fraction E P subscript A equals 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 0 comma 05 end fraction plus fraction numerator 9 cross times 10 to the power of 9 cross times 10 to the power of negative 6 end exponent cross times 3 cross times 10 to the power of negative 6 end exponent over denominator 0 comma 05 end fraction E P subscript A equals open parentheses 360 cross times 10 to the power of negative 3 end exponent close parentheses plus open parentheses 540 cross times 10 to the power of negative 3 end exponent close parentheses E P subscript A equals 900 cross times 10 to the power of negative 3 end exponent E P subscript A equals 0 comma 9 space straight J
     
  2. Menentukan energi potensial di titik B

    E P subscript B equals E P subscript B A end subscript plus E P subscript B C end subscript E P subscript B equals fraction numerator k q subscript B q subscript A over denominator r subscript B A end subscript end fraction plus fraction numerator k q subscript B q subscript C over denominator r subscript B C end subscript end fraction E P subscript B equals fraction numerator 9 cross times 10 to the power of 9 cross times 2 cross times 10 to the power of negative 6 end exponent cross times 10 to the power of negative 6 end exponent over denominator 0 comma 05 end fraction plus fraction numerator 9 cross times 10 to the power of 9 cross times 2 cross times 10 to the power of negative 6 end exponent cross times 3 cross times 10 to the power of negative 6 end exponent over denominator 0 comma 05 end fraction E P subscript B equals open parentheses 360 cross times 10 to the power of negative 3 end exponent close parentheses plus open parentheses 1080 cross times 10 to the power of negative 3 end exponent close parentheses E P subscript B equals 1440 cross times 10 to the power of negative 3 end exponent E P subscript B equals 1 comma 44 space straight J
     
  3. Menentukan energi potensial di titik C

    E P subscript C equals E P subscript C A end subscript plus E P subscript C B end subscript E P subscript C equals fraction numerator k q subscript C q subscript A over denominator r subscript C A end subscript end fraction plus fraction numerator k q subscript C q subscript B over denominator r subscript C B end subscript end fraction E P subscript C equals fraction numerator 9 cross times 10 to the power of 9 cross times 3 cross times 10 to the power of negative 6 end exponent cross times 10 to the power of negative 6 end exponent over denominator 0 comma 05 end fraction plus fraction numerator 9 cross times 10 to the power of 9 cross times 3 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 0 comma 05 end fraction E P subscript C equals open parentheses 540 cross times 10 to the power of negative 3 end exponent close parentheses plus open parentheses 1080 cross times 10 to the power of negative 3 end exponent close parentheses E P subscript C equals 1620 cross times 10 to the power of negative 3 end exponent E P subscript C equals 1 comma 62 space straight J
     
  4. Menentukan energi potensial sistem

    E P subscript S equals E P subscript A B end subscript plus E P subscript B C end subscript plus E P subscript C A end subscript E P subscript S equals fraction numerator k q subscript A q subscript B over denominator r subscript A B end subscript end fraction plus fraction numerator k q subscript B q subscript C over denominator r subscript B C end subscript end fraction plus fraction numerator k q subscript C q subscript A over denominator r subscript C A end subscript end fraction E P subscript S equals 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 0 comma 05 end fraction plus fraction numerator 9 cross times 10 to the power of 9 cross times 2 cross times 10 to the power of negative 6 end exponent cross times 3 cross times 10 to the power of negative 6 end exponent over denominator 0 comma 05 end fraction plus fraction numerator 9 cross times 10 to the power of 9 cross times 3 cross times 10 to the power of negative 6 end exponent cross times 10 to the power of negative 6 end exponent over denominator 0 comma 05 end fraction E P subscript S equals open parentheses 360 cross times 10 to the power of negative 3 end exponent close parentheses plus open parentheses 1080 cross times 10 to the power of negative 3 end exponent close parentheses plus open parentheses 540 cross times 10 to the power of negative 3 end exponent close parentheses E P subscript S equals 1980 cross times 10 to the power of negative 3 end exponent E P subscript S equals 1 comma 98 space straight J
     

Jadi, energi potesial pada titik A, B, C, dan energi potensial sistem berturut-turut adalah 0,9 J, 1,44 J, 1,62 J, dan 1,98 J.space space

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