Electric charge on the plastic cube: [tex]1.3\cdot 10^{-7}C[/tex]
Explanation:
The electric potential around a charged sphere (such as the Van der Graaf) generator is given by
[tex]V(r)=\frac{kQ}{r}[/tex]
where
k is the Coulomb's constant
Q is the charge on the sphere
r is the distance from the centre of the sphere
Here we have:
V = 200,000 V on the surface of the sphere, so at r = 12.0 cm
We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at
r' = 12.0 + 2.0 = 14.0 cm
Since the voltage is inversely proportional to r, we can use:
[tex]Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V[/tex]
This is the potential at the location of the plastic cube.
Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:
[tex]qV' = \frac{1}{2}mv^2[/tex]
where:
q is the charge on the plastic cube
V' is the potential at the location of the cube
m = 5.0 g = 0.005 kg is the mass of the cube
v = 3.0 m/s is the final speed of the cube
Solving for q, we find the charge on the cube:
[tex]q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C[/tex]
Learn more about electric fields:
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