The belt of an electrostatic generator is 50cm wide and travels 30cm/s. The belt carries charge into the sphere at a rate corresponding to 10−4A. What is the surface density of charge on the belt?
A
6.7×10−5Cm−2
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B
6.7×10−7Cm−2
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C
6.7×10−4Cm−2
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D
6.7×10−8Cm−2
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Solution
The correct option is C6.7×10−4Cm−2 Let the surface charge density of the belt be σ.
Velocity of the belt, V=30cm/s=0.3m/s
Width of the belt, w=50cm=0.5m
Since surface charge density, σ=QA Q=σ×A
Dividing both sides by time t Qt=σ×At(1)
Since Qt=I and A=l×w
where l is the length of the belt ⇒At=l×wt=V×w
Therefore equation (1) becomes, I=σ×V×w 10−4=σ×0.3×0.5 ⇒σ=6.67×10−4Cm−2≈6.7×10−4Cm−2