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A particle having a charge of q=8.85muC ...

A particle having a charge of `q=8.85muC` is placed on the axis of a circular ring of radius `R=30cm` at a point `P` at a distance of `a=40cm` from the centre of the ring.The electric flux passing through the ring is `x xx 10^(5)N//C` .Find the value of `x`?

Text Solution

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The correct Answer is:
1
`phi_(E)= phi_(E)=(q)/(2epsilon_(0))(1-(a)/(sqrt(R^(2)+a^(2))))`
` rArr phi_(E)=(8.85xx10^(-6))/(2(8.85xx10^(-12)))(1-(40)/(50))`
`rArr phi_(E)=(10^(6))/(2)(1-(4)/(5))`
`rArr phi_(E)=1xx10^(5)=x xx 10^(5)x=1`
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