A 0.22 kg air track glider moving at 0.60 m/s a collides elastically into a 0.44 kg glider at rest. After collision the 0.22 kg
1 answer:
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Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:
ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:
r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:
Finally, you obtain for E:
hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C