qP = +6 × 10¯⁷ C
qQ = -8 × 10¯⁸ C
jarak (r) = 2 cm = 2 × 10¯² m
k = 9 x 10⁹ Nm²/C²
gaya (F)...?
F = (9 x 10⁹ Nm²/C² x 6 × 10¯⁷ C x -8 × 10¯⁸ C)/(2 × 10¯² m)²
= (9 x 6 x -8 x 10¯⁶) /(4 × 10¯⁴)
= -108 x 10¯² N
= -1,08 N (negatif menunjukkan gaya tarik menarik)
Jadi, besar gaya tarik menarik kedua muatan adl 1,08 N.
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
diket:
qP = +6 × 10¯⁷ C
qQ = -8 × 10¯⁸ C
jarak (r) = 2 cm = 2 × 10¯² m
k = 9 x 10⁹ Nm²/C²
ditanya:
gaya (F)...?
jawab:
F = (k qP qQ)/r²
F = (9 x 10⁹ Nm²/C² x 6 × 10¯⁷ C x -8 × 10¯⁸ C)/(2 × 10¯² m)²
= (9 x 6 x -8 x 10¯⁶) /(4 × 10¯⁴)
= -108 x 10¯² N
= -1,08 N (negatif menunjukkan gaya tarik menarik)
Jadi, besar gaya tarik menarik kedua muatan adl 1,08 N.