Question

7. I] Two steel balls, each 1.00 kg mass, and each holding +1.00 С charge, are separated 1.00 meters apart. (a) The gravitational force between them is (I) attractive or (2) repulsive, and of what magnitude? tb) The electrical forc between hem is () atractive or (a) repalive, and or what magnituet

Answer 1

Part A.

Remember that gravitational force is given by:

Fg = G*m1*m2/R^2

Also Gravitational force between two object will always be attractive.

G = Gravitational constant = 6.67*10^-11

m1 = m2 = 1.00 kg

R = 1.00 m

So,

Fg = 6.67*10^-11*1.00*1.00/1.00^2

Fg = 6.67*10^-11 N = Magnitude of gravitational force

Part B.

Electrostatic force is given by:

Fe = k*q1*q2/R^2

Now electrostatic force between two charge with same sign will always be repulsive and force between two charges with different sign will always be attractive.

So in this case force is Repulsive.

k = electrostatic constant = 9*10^9

q1 = q2 = 1.00 C

R = 1.00 m

So,

Fe = 9*10^9*1.00*1.00/1.00^2

Fe = 9*10^9 N = Magnitude of electrostatic force

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