Question

please bantu...besok udh harus dikumpulkan ​

Answer 1

Jawaban:

★F1: F antara 2Q dan Q

F1 =k×Q×2Q/l²

F1 =2 kQ/l²

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★F2 F antara 2Q dan 4Q (jarak = √2 l )

F2=k×2Q×4Q/ (√2 l )²

F2=kQ²/l²

F2x = F2 cos 45°

F2x =4kQ²/l²×½√2

F2x =2√2kQ²/l²

F2y= F2 sin 45°

F2y=4kQ²/l²×½√2

F2y=2√2kQ²/l²

★F3 F antara 2Q dan 3Q

F3=k×Q×3Q/l²

F3=6kQ²/l²

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★ΣFx = F1+ F2x

ΣFx = 2kQ²/l² +2√2kQ²/l²

ΣFx =(2+2√2) (9+10⁹)Q²/l²

ΣFx =4.35 ×10¹⁰Q²/l² N

ΣFy = F3 + F2y

ΣFy =6kQ²/l² + 2√2kQ²/l²

ΣFy =(6+2√2) (9+10⁹)Q²/l²

ΣFy =7.95 ×10¹⁰Q²/l² N

F = √(ΣFx² +ΣFy²)

F = 9.06 × 10¹⁰Q²/l² N

arah F = a tan (Fy/ZFx)

arah F = a tan (7,95/4,35)

arah F = a tan 1,83

arah F = 61,31°

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