Answer:
[tex]8.33\times 10^{-6}\ C[/tex]
3333333.33 V/m
Explanation:
k = Coulomb constant = [tex]9\times 10^9\ Nm^2/C^2[/tex]
d = Diameter = 30 cm
r = Radius = [tex]\frac{d}{2}=\frac{30}{2}=15\ cm[/tex]
V = Voltage = 500000 V
Potential in a Van de Graaff generator is given by
[tex]V=\frac{kq}{r}\\\Rightarrow q=\frac{Vr}{k}\\\Rightarrow q=\frac{500000\times 0.15}{9\times 10^9}\\\Rightarrow q=8.33\times 10^{-6}\ C[/tex]
Charge needed on the sphere is [tex]8.33\times 10^{-6}\ C[/tex]
Electric field is given by
[tex]E=\frac{V}{r}\\\Rightarrow E=\frac{500000}{0.15}\\\Rightarrow E=3333333.33\ V/m[/tex]
The electric field strength just outside the surface of the sphere is 3333333.33 V/m
(a). The value of the charge on the sphere is [tex]8.34*10^{-6}C[/tex]
(b). The electric field strength just outside the surface of the sphere is [tex]3.34*10^{6}V/m[/tex]
It is computed as,
[tex]V=k\frac{q}{r}[/tex]
Where,
Given that, [tex]V=500000,r=30/2=15cm=0.15m[/tex]
Substitute values in above formula.
[tex]500000=9*10^{9}*\frac{q}{0.15} \\\\q=500000*\frac{0.15}{9*10^{9} } \\\\q=8.34*10^{-6}C[/tex]
As we know that,
Electric field,[tex]E=\frac{V}{r}[/tex]
[tex]E=\frac{500000}{0.15} \\\\E=3.34*10^{6}V/m[/tex]
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