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Sebuah bujur sangkar ABCD dengan rusuk 20 cm seperti gambar. Hitung besar kuat medan listrik di titik D jika qA = qC = + dan qB = - ....

Pertanyaan

Sebuah bujur sangkar ABCD dengan rusuk 20 cm seperti gambar.

Hitung besar kuat medan listrik di titik D jika qA = qC = +4 space μC dan qB = -4 space μC ....

Pembahasan Video:

Pembahasan Soal:

Diketahui

s equals 20 space cm equals 0 comma 2 space straight m q subscript A equals q subscript C equals plus 4 space μC equals plus 4 cross times 10 to the power of 6 space straight C q subscript B equals negative 4 space μC equals negative 4 cross times 10 to the power of 6 space straight C

Ditanya :

E subscript D equals... ?

Penyelesaian:

Langkah pertama, gambarkan arah medan listrik di titik D akibat masing-masing muatan yang terletak di titik A, B, dan C. 

Hitung besar medan listrik EA, EB, dan EC

Menghitung medan listrik di titik D akibat muatan A (EA):

E subscript A equals k times q subscript A over r subscript A D end subscript squared E subscript A equals 9 cross times 10 to the power of 9 times fraction numerator 4 cross times 10 to the power of negative 6 end exponent over denominator 0 comma 2 squared end fraction E subscript A equals fraction numerator 36 cross times 10 cubed over denominator 0 comma 04 end fraction E subscript A equals 900 cross times 10 cubed E subscript A equals 9 cross times 10 squared cross times 10 cubed E subscript A equals 9 cross times 10 to the power of 5 space straight N divided by straight C 

Menghitung medan listrik di titik D akibat muatan C (EC):

table attributes columnalign right center left columnspacing 0px end attributes row cell E subscript C end cell equals cell k times q subscript C over r subscript C D end subscript squared end cell row cell E subscript C end cell equals cell 9 cross times 10 to the power of 9 times fraction numerator 4 cross times 10 to the power of negative 6 end exponent over denominator 0 comma 2 squared end fraction end cell row cell E subscript C end cell equals cell fraction numerator 36 cross times 10 cubed over denominator 0 comma 04 end fraction end cell row cell E subscript C end cell equals cell 900 cross times 10 cubed end cell row cell E subscript C end cell equals cell 9 cross times 10 squared cross times 10 cubed end cell row cell E subscript C end cell equals cell 9 cross times 10 to the power of 5 space straight N divided by straight C end cell end table 

Menghitung medan listrik di titik D akibat muatan B (EB):

E subscript B equals k times q subscript B over r subscript B D end subscript squared

terlebih dahulu menghitung rBD2

r subscript B D end subscript squared equals s squared plus s squared r subscript B D end subscript squared equals 2 times s squared r subscript B D end subscript squared equals 2 times 0 comma 2 squared r subscript B D end subscript squared equals 2 times 0 comma 04 r subscript B D end subscript squared equals 0 comma 08 

Sehingga, 

table attributes columnalign right center left columnspacing 0px end attributes row cell E subscript B end cell equals cell k times q subscript B over r subscript B D end subscript squared end cell row cell E subscript B end cell equals cell 9 cross times 10 to the power of 9 times fraction numerator 4 cross times 10 to the power of negative 6 end exponent over denominator 0 comma 08 end fraction end cell row cell E subscript B end cell equals cell fraction numerator 36 cross times 10 cubed over denominator 0 comma 08 end fraction end cell row cell E subscript B end cell equals cell 450 cross times 10 cubed end cell row cell E subscript B end cell equals cell 4 comma 50 cross times 10 squared cross times 10 cubed end cell row cell E subscript B end cell equals cell 4 comma 50 cross times 10 to the power of 5 space straight N divided by straight C end cell end table 

Selanjutnya, menghitung resultan medan listrik di titik D dengan menggunakan metode penguraian vektor, yaitu:

Kemudian menghitung EBx dan EBy 

table attributes columnalign right center left columnspacing 0px end attributes row cell E subscript B x end subscript over E subscript B end cell equals cell fraction numerator s over denominator s square root of 2 end fraction end cell row cell fraction numerator E subscript B x end subscript over denominator 4 comma 5 cross times 10 to the power of 5 end fraction end cell equals cell fraction numerator 1 over denominator square root of 2 end fraction end cell row cell E subscript B x end subscript end cell equals cell fraction numerator 4 comma 5 cross times 10 to the power of 5 over denominator square root of 2 end fraction end cell row cell E subscript B x end subscript end cell equals cell fraction numerator 4 comma 5 cross times 10 to the power of 5 over denominator 1 comma 414 end fraction end cell row cell E subscript B x end subscript end cell equals cell 3 comma 18 cross times 10 to the power of 5 space straight N divided by straight C end cell end table 

E subscript B y end subscript equals E subscript B x end subscript equals 3 comma 18 cross times 10 to the power of 5 space straight N divided by straight C

Selanjutnya menghitung resultan vektor medan listrik pada arah sumbu-x dan sumbu-y:

capital sigma E subscript x equals E subscript B x end subscript minus E subscript C capital sigma E subscript x equals 3 comma 18 cross times 10 to the power of 5 minus 9 cross times 10 to the power of 5 capital sigma E subscript x equals negative 5 comma 82 cross times 10 to the power of 5 space straight N divided by straight C  capital sigma E subscript y equals E subscript A minus E subscript B y end subscript capital sigma E subscript x equals 9 cross times 10 to the power of 5 minus 3 comma 18 cross times 10 to the power of 5 capital sigma E subscript x equals 5 comma 82 cross times 10 to the power of 5 space straight N divided by straight C

Setelah itu menghitung resultan medan listrik total di titik D, yaitu:

E subscript D equals square root of open parentheses capital sigma E subscript x close parentheses squared plus open parentheses capital sigma E subscript y close parentheses squared end root E subscript D equals square root of open parentheses negative 5 comma 82 cross times 10 to the power of 5 close parentheses squared plus open parentheses 5 comma 82 cross times 10 to the power of 5 close parentheses squared end root E subscript D equals square root of open parentheses 5 comma 82 cross times 10 to the power of 5 close parentheses squared plus open parentheses 5 comma 82 cross times 10 to the power of 5 close parentheses squared end root E subscript D equals square root of 2 times open parentheses 5 comma 82 cross times 10 to the power of 5 close parentheses squared end root E subscript D equals 5 comma 82 cross times 10 to the power of 5 square root of 2 E subscript D equals 5 comma 82 square root of 2 cross times 10 to the power of 5 space straight N divided by straight C 

Jadi, kuat medan listrik di titik D adalah 5 comma 82 square root of 2 cross times 10 to the power of 5 space straight N divided by straight C

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 30 Agustus 2021

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Dua muatan listrik muatannya 0,2 C diletakkan di titik B dan C sebuah segitiga sama kaki. Jika, maka resultan kuat medan listrik di titik A adalah....

Pembahasan Soal:

Kuat medan listrik yang disebabkan oleh muatan uji dirumuskan dengan persamaan :

E equals fraction numerator k q over denominator r squared end fraction

Maka :

E subscript B A end subscript equals fraction numerator k q subscript B over denominator r squared subscript B A end subscript end fraction  E subscript B A end subscript equals fraction numerator left parenthesis 9 cross times 10 to the power of 9 right parenthesis left parenthesis 0 comma 2 cross times 10 to the power of negative 6 end exponent right parenthesis over denominator left parenthesis 0 comma 3 right parenthesis squared end fraction  E subscript B A equals 2 cross times 10 to the power of 4 N divided by C

Lalu :

E subscript C A end subscript equals fraction numerator k q subscript c over denominator r squared subscript c A end subscript end fraction  E subscript C A end subscript equals fraction numerator left parenthesis 9 cross times 10 to the power of 9 right parenthesis left parenthesis 0 comma 2 cross times 10 to the power of negative 6 end exponent right parenthesis over denominator left parenthesis 0 comma 3 right parenthesis squared end fraction  E subscript C A end subscript equals 2 cross times 10 to the power of 4 N divided by C

Misalkan :

E subscript B A end subscript equals E subscript C A end subscript equals E equals 2 cross times 10 to the power of 4 space N divided by C

Sehingga :

E subscript A equals square root of E subscript B A end subscript superscript 2 plus E subscript C A end subscript superscript 2 plus 2 E subscript B A end subscript E subscript C A end subscript cos theta end root  E subscript A equals square root of E squared plus E squared plus 2. E. E left parenthesis 0 right parenthesis end root  E subscript A equals square root of 2 E squared end root  E subscript A equals E square root of 2

Jadi :

E subscript A equals 2 square root of 2 cross times 10 to the power of 4 space N divided by C

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Pada dua titik sudut segitiga sama sisi dengan panjang sisi 30 cm diletakkan dua muatan listrik yang besarnya sama yaitu 0,9 µC, maka resultan kuat medan listrik pada titik sudut ketiga adalah ....

Pembahasan Soal:

Karena besar muatan dan jarak dari muatan ke titik sudut ketiga, maka E1 = E2 = E

table attributes columnalign right center left columnspacing 0px end attributes row E equals cell fraction numerator 9 cross times 10 to the power of 9 times 9 cross times 10 to the power of negative 7 end exponent over denominator 0 comma 3 squared end fraction end cell row E equals cell 9 cross times 10 to the power of 4 space straight N divided by straight C end cell row cell E subscript t o t end subscript end cell equals cell square root of E squared plus E squared plus 2 E times E space cos space 60 to the power of 0 end root end cell row cell E subscript t o t end subscript end cell equals cell square root of E squared plus E squared plus 2 E squared open parentheses 1 half close parentheses end root end cell row cell E subscript t o t end subscript end cell equals cell square root of 3 E squared end root end cell row cell E subscript t o t end subscript end cell equals cell E square root of 3 end cell row cell E subscript t o t end subscript end cell equals cell 9 cross times 10 to the power of 4 square root of 3 space straight N divided by straight C end cell end table

Jadi, jawaban yang tepat adalah E

0

Roboguru

Dua buah muatan identik q  masing-masing diletakkan pada koordinat (r,0)  dan (0,r). Resultan kuat medan listrik di pusat koordinat adalah ….

Pembahasan Soal:

Gambarkan medan-medan yang muncul di pusat koordinat.

begin mathsize 14px style rightwards double arrow E subscript 0 equals square root of E subscript 1 superscript 2 plus E subscript 2 superscript 2 end root rightwards double arrow E subscript 0 equals square root of open parentheses k q over r squared close parentheses squared plus open parentheses k q over r squared close parentheses squared end root rightwards double arrow E subscript 0 equals square root of 2 k squared q squared over r to the power of 4 end root rightwards double arrow E subscript 0 equals k q over r squared square root of 2 end style  

Dengan demikian jawaban yang tepat adalah A.

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Roboguru

Dua bola bermuatan listrik B dan C berada pada jarak 0,5 m. Muatan bola B dan C masing-masing besarnya qB​=+40×10−5Cdan qC​=−50×10−5C. Titik A berada dalam pengaruh medan listrik dari kedua bola terse...

Pembahasan Soal:

Diketahui :

qB=+40×105Cq=50×105Cθ=45rAC=1m

Ditanya : kuat medan listrik di A?

Penyelesaian :

Karena titik A, B dan C membentuk segitiga sama kaki, maka jarak muatan B ke titik A adalah 1 m

Medan listrik dari muatan B berarah menjauhi muatan B

EB=rAB2kqBEB=129×10940×105EB=3,6×106N/C

Medan listrik dari muatan C menuju muatan C

EC=rAB2kqBEC=129×10950×105EC=4,5×106N/C

maka, medan listrik di A adalah

EA=EB2+EC2+2EBECcos(18045)EA=(3,6×106)2+(4,5×106)2+23,6×1064,5×106cos135EA=1,296×1013+2,025×10132,23×1013EA=3,3×106N/C

Dengan demikian, kuat medan listrik di A adalah 3,3×106N/C.

0

Roboguru

Tiga buah muatan diletakkan pada ketiga sudut suatu persegi seperti gambar. Setiap sisi dari persegi adalah 30 cm. Hitung besarnya E di sudut keempatnya!

Pembahasan Soal:

Diketahui

r equals 30 space cm equals 0 comma 3 space straight m q subscript 1 equals 4 space μC equals 4 cross times 10 to the power of negative 6 end exponent space straight C q subscript 2 equals 5 space μC equals 5 cross times 10 to the power of negative 6 end exponent space straight C q subscript 3 equals 8 space μC equals 8 cross times 10 to the power of negative 6 end exponent space straight C

Ditanyakan ER

Penyelesaian

Tentukan besar jarak dari muatan q3 ke sudut keempatnya

r subscript 3 equals square root of r squared plus r squared end root r subscript 3 equals square root of 0 comma 3 squared plus 0 comma 3 squared end root r subscript 3 equals 0 comma 3 square root of 2 space straight m

Tentukan besar medan dari E4, E5 dan E8

E4=r2kq1E4=(0,3)29×1094×106E4=4×105N/CE5=r2kq2E5=(0,3)29×1095×106E5=5×105N/CE8=r2kq3E8=(0,32)29×1098×106E8=4×105N/C

Hitung resultan dari E4 dan E5

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Medan E6 dan E8 saling berlawanan arah maka Resultan totalnya adalah

E equals E subscript 8 minus E subscript 6 E equals 4 cross times 10 to the power of 5 minus 6 comma 4 cross times 10 to the power of 5 E equals negative 2 comma 4 cross times 10 to the power of 5 space straight N divided by straight C

Dengan demikian besarnya E di sudut keempatnya adalah bold 2 bold comma bold 4 bold cross times bold 10 to the power of bold 5 bold space bold N bold divided by bold C

 

 

0

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