Fisika

Pertanyaan

Mohon bantuannya ya..
Pake cara. Kalo asal aku teport
Mohon bantuannya ya.. Pake cara. Kalo asal aku teport

1 Jawaban

  • Listrik Statis

    [tex] Q_A = 6 \ \mu C = 6 \times 10^{-6} C \\
    Q_C = - 2 \ \mu C = -2 \times 10^{-6} C \\
    r_{AB} = 2 \ cm = 2 \times 10^{-2} m \\
    r_{BC} = 1 \ cm = 10^{-2} m \\
    \\
    a. V_{BA} = \frac{k \ Q_A}{r_{AB}} \\
    V_{BA} = \frac{9 \times 10^9 \cdot 6 \times 10^{-6} } { 2 \times 10^{-2} } \\
    V_{BA} = 2,7 \times 10^6 \ V
    \\ \\
    b. V_{BC} = \frac{k \ Q_C}{r_{BC}} \\
    = \frac{9 \times 10^9 \cdot (-2 \times 10^{-6} ) } {10^{-2} } \\
    V_{BA} = 1,8 \times 10^6 \ V \\ \\
    c. E_{BA} = \frac{k \ Q_A}{(r_{AB})^2} \\
    E_{BA} = \frac{9 \times 10^9 \cdot 6 \times 10^{-6} }{ (2 \times 10^{-2} )^2 } \\
    E_{BA} = 1,35 \times 10^8 \ N/C \\
    \\
    E_{BC} = \frac{k \ Q_C}{(r_{AB})^2} \\
    E_{BC} = \frac{9 \times 10^9 \cdot 2 \times 10^{-6} }{ (10^{-2} )^2 } \\
    E_{BC} = 1,8 \times 10^8 \ N/C \\
    \\
    Maka : \\
    E_B = E_{BC} - E_{AB} \\
    = 1,8 \times 10^8 - 1,35 \times 10^8
    = 4,5 \times 10^7 \ N/C [/tex]