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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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irina irina
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Posts: 919
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11 years ago
An isolated aluminum sphere of radius 2.5 cm is at a potential of 400 V. How many electrons have been removed from the sphere to raise it to this potential?
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wrote...
#1
11 years ago
Charge q = Vr/k, where k = 8.99E9 Nm^2/C^2.
The  number of electrons = q*6.2415E18.
obb
wrote...
#2
11 years ago
v=400 V
r=0.025 m
k=9*10^9

V=K*q/r
sovle for q

q=(v*r)/k
(400*0.025)/(9*10^9) = +1.11111*10^-9 C

So, a charge of +1.11111*10^-9 C has been removed

The charge of one electron is:
e= -1.602*10^-19 C

Number of electrons removed
=q/e
1.11111*10^-9/(1.602*10^-19) = 6,935,767,790 electrons
"rounded"
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