(a) Two thin hollow plastic spheres, about the size of a ping-pong ball with mas
ID: 2228783 • Letter: #
Question
(a) Two thin hollow plastic spheres, about the size of a ping-pong ball with masses (m1 = m2 = 2e-3 kg) have been rubbed with wool. Sphere 1 has a charge q1 = -4e-9 C and is at location m. Sphere 2 has a charge q2 = -4e-9 C and is at location m. It will be useful to draw a diagram of the situation, including the relevant vectors. What is the relative position vector pointing from q1 to q2? What is the distance between q1 and q2? What is the unit vector in the direction of ? What is the magnitude of the gravitational force exerted on q2 by q1? What is the gravitational force (vector) exerted on q2 by q1? What is the electric force (vector) exerted on q2 by q1? What is the ratio of the magnitude of the electric force to the magnitude of the gravitational force? (You see that electric forces between two small charged objects are typically very much larger than gravitational forces between those same small objects. It takes the entire mass of the Earth to exert a sizable gravitational force on a small object.) If the two masses were 5 times farther away (that is, if the distance between the masses were 5 times ), what would be the ratio of the magnitude of the electric force to the magnitude of the gravitational force now?Explanation / Answer
As gravitational constant G=6.67398 × 10^-11
electric constant K=8.85418782 × 10^-12
r=sqrt((-70X10^-2 x + 60X10^-2 y + 0 z)/0.9219544) unit vector
Magnitude of Fgrav=6.67398X 10^-11 (0.000004)/(0.9219544)=24.5X10^-17N
Fgra=24.5X10^-17 r N
Felectric= 8.85418782 × 10^-12(16X10^-18)/(0.9219544)=153.6X10^-30 N
Felectric=153.6X10^-30 r N
Felectric/Fgrav =6.27X10^-13
If the masses are five times away, then their ratio will not change because icreasing five in both forces will cancell out.
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