Homework help:A small plastic sphere with a mass of 2.60 X 10-15 kg

AI Thread Summary
The discussion centers on a small plastic sphere with a mass of 2.60 x 10^-15 kg suspended between two charged plates with a potential difference of 265.4 V. The sphere is determined to have a positive charge because it is attracted to the negatively charged upper plate, ensuring a net force of zero. The magnitude of the sphere's charge is calculated to be 4.8 x 10^-19 C, leading to a deficit of 3.0 electrons. The calculations were confirmed to be correct after addressing an initial error in the charge's power. The final conclusion reaffirms the sphere's positive charge and the number of excess or deficit electrons.
Jaimie
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Homework Statement



"A small plastic sphere (with a mass of 2.60 X 10-15 kg) is suspended between
two plates, with a potential difference of 265.4 V across the plates. The separation of the plates is
0.500 cm. The upper plate is negative and the lower one is positive.
a)What type of charge is on the sphere?Explain your reasoning.
b)What is the magnitude of the charge on the sphere?
c)How many excess or deficit electrons does the sphere have?"

Homework Equations



q=mgd/∆V
q=Ne

The Attempt at a Solution



a) The charge on the sphere is positive. In order for the sphere to suspend, there must be a net force of zero on the object, where the magnitude of the upwards electrical force is equal to the downwards gravitational force. As the electrical force is pointed upwards (acting in the direction of the electrical field), it must be attracted to the negatively-charged plate. If the sphere’s charge were negative, the electrical force would be repelling from the negative plate and pointed downwards, where the net force acting on the object would no longer be zero.

b) Fe= εq; ε= ∆V/d
Fe= Fg
(∆V/d)(q)= mg
∴q=mgd/∆V=2.60 x 10^-15 kg(9.8N/kg)(0.500cm)(1m/100cm)/265.4V=4.8 x 10^-4 C

c) q=Ne
q/e=N
∴N= q/e= 4.8 x 10^-4 C /(1.6 x 10^-19 C) =3.0 x 10^15 electrons
∴Deficit of 3.0 x 10^15 electrons, as charge is positive.

Please advise if this is correct?
 
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Jaimie said:
(∆V/d)(q)= mg
∴q=mgd/∆V=2.60 x 10^-15 kg(9.8N/kg)(0.500cm)(1m/100cm)/265.4V=4.8 x 10^-4 C

Your method looks good. But check to see if you really get a power of -4 in the answer for the charge.
 
Oh...calculation error...this should be 4.8 x 10^-19 C. TSny, thank you for letting me know.
 
So then for c)
q=Ne
q/e=N
∴N= q/e= 4.8 x 10^-19 C /(1.6 x 10^-19 C) =3.0 electrons
∴Deficit of 3.0 electrons, as charge is positive.
 
That looks good.
 
Thank you for your help!
 
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