contestada

There are two identical small metal spheres
with charges 22.8 µC and −16.7668 µC. The
distance between them is 6 cm. The spheres
are placed in contact then set at their original
distance.
Calculate the magnitude of the force between the two spheres at the final position. The value of the Coulomb constant
is 8.98755 × 10^9
Answer in units of N.

Respuesta :

olatunbosunadeoke

Answer: -954.4N

Explanation:

Given that,

Positive charge (q1) = 22.8 µC

22.8 µC = 22.8 x 10^-6C

Negative charge (q2) = −16.7668 µC

−16.7668 µC = −16.7668 x 10^-6C

Distance between charges (r) = 6cm

convert centimetres to metres

100cm = 1m

6cm = (6/100) = 0.06m

force between the two charges (F) = ?

Recall that k is Coulomb's constant with a value of 8.98755 × 10^9 Nm2C−2

So, apply the formula for F= (kq1q2) / r²

F = (8.98755 × 10^9 Nm2C−2 x 22.8 x 10^-6C x −16.7668 x 10^-6C) / (0.06m)²

F = (-3440.55 x 10^-3Nm²) / 0.0036m²

F =(-3440.55 x 10^-3Nm²) / 3.6 x 10^-3m²

F = -954.4N

Thus, the magnitude of the force between the two spheres is -954.4 Newton.

      The correct Answer is: -954.4N

  • First Given that is:
  • When in a Positive charge (q1) = 22.8 µC
  • so that 22.8 µC = 22.8 x 10^-6C
  • Then Negative charge (q2) = −16.7668 µC
  • After that −16.7668 µC = −16.7668 x 10^-6C
  • Then Distance between the charge is  (r) = 6cm
  • It convert the centi-meters to meters
  • 100cm = 1m
  • 6cm = (6/100) = 0.06m
  • When force between the two charges (F) = ?
  • Then Recall that k is Coulomb's constant with a value of 8.98755 × 10^9 Nm2C−2  

          So, that apply the formula for F= (kq1q2) / r²

  1. F = (8.98755 × 10^9 Nm2C−2 x 22.8 x 10^-6C x −16.7668 x 10^-6C) /(0.06m)²
  2. F = (-3440.55 x 10^-3Nm²) / 0.0036m²
  3. F =(-3440.55 x 10^-3Nm²) / 3.6 x 10^-3m²
  4. F = -954.4N  
  • Thus, therefore after that, the magnitude of the force between the two spheres is -954.4 Newton.

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