Diketahui:
• E = 40 N/C
• Q = 2 cm = 0,02 m
• k = 9 x 10⁹ Nm²/C²
ㅤㅤㅤㅤㅤㅤ
Ditanyakan:
Berapakah besar muatan Q ?
Penyelesaian:
[tex] \sf E = k \: \dfrac{Q}{ {r}^{2} } [/tex]
[tex] \sf Q = \dfrac{E \: {r}^{2} }{k } [/tex]
[tex] \sf Q = \dfrac{40 \: N/C \times {(0,02 \: m)}^{2}}{9 \times 1 {0}^{9} \: N {m}^{2} /C^{2} } [/tex]
[tex] \sf Q = 1,78 \times {10}^{ - 12} \: C[/tex]
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Diketahui:
• E = 40 N/C
• Q = 2 cm = 0,02 m
• k = 9 x 10⁹ Nm²/C²
ㅤㅤㅤㅤㅤㅤ
Ditanyakan:
Berapakah besar muatan Q ?
ㅤㅤㅤㅤㅤㅤ
Penyelesaian:
[tex] \sf E = k \: \dfrac{Q}{ {r}^{2} } [/tex]
[tex] \sf Q = \dfrac{E \: {r}^{2} }{k } [/tex]
[tex] \sf Q = \dfrac{40 \: N/C \times {(0,02 \: m)}^{2}}{9 \times 1 {0}^{9} \: N {m}^{2} /C^{2} } [/tex]
[tex] \sf Q = 1,78 \times {10}^{ - 12} \: C[/tex]