Two small plastic spheres are given positive electrical charges. When they are a

Question

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8 cm apart, the repulsive force between them has a magnitude of 0.250 N .

Part A

What is the charge on each sphere if the two charges are equal?

5.73•1013c

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Part B

What is the charge on first sphere if it has four times the charge of the other?

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Part C

What is the charge on the second sphere?

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8 cm apart, the repulsive force between them has a magnitude of 0.250 N .

Part A

What is the charge on each sphere if the two charges are equal?

Q=

5.73•1013c

  C  

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Incorrect; Try Again; 5 attempts remaining

Part B

What is the charge on first sphere if it has four times the charge of the other?

Q1= C

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Part C

What is the charge on the second sphere?

Explanation / Answer

One Charge = q1 = q

Force = F = 0.250 N

Other charge = q2 =q

Distance = r = 14.8 cm = 0.148 m

permittivity of free space = eo =8.854×1012 C^2/ (N.m^2)

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q = sq rt [ 0.250[4pi8.85×10^12 ] (0.148*0.148)

q =0.78*10^-6C

Each charge is 7.8*10^-7C
______________________________________...

One Charge = q1 = q

Force = F = 0.250 N

Other charge = q2 =4q1 = 4q

Distance = r = 0.148 m

permittivity of free space = eo =8.854×1012 C^2/ (N.m^2)

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

q1q2 =F [4pieo]r^2

(q)(4q) = F[4pieo]r^2

q = sq rt { 0.250[4pi8.85×10^12 ] (0.148*0.148)/ 4 }

q = 0.39*10^-6 C

Edited

Therefore,charge on first sphere =4q =4*3.9*10^-7 C=1.56*10^-6 C

If the charge on first sphere is four times the charge of the other, then charge on first sphere is 1.56*10^-6 C and charge on the second sphere is 3.9*10^-7 C _________________________________-

C)The charge on the second sphere is 3.9*10^-7 C

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