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a unifrom surface charge density 8sigma ...

a unifrom surface charge density `8sigma` exists over the entire `x-y` plane except for a circular hole of radius of a point `p(0,0,sqrt(3))` on the z-axis is found to be `2sqrt(x)(sigma)/(epsilon_(0))` ,Find the value of `x`.

Text Solution

Verified by Experts

The correct Answer is:
3

`E_(net)=(sigma)/(2epsilon_(0))-E_(dis)`
`rArr E_(net)=(sigma')/(2epsilon_(0))-(sigma)/(2epsilon_(0))(1-(z)/(sqrt(z^(2)+a^(2))))`
`E_("net")=(sigma')/(2epsilon_(0)sqrt(z^(2)+a^(2)))`
` rArr E_("net")=((8sigma)sqrt(3)a)/(2epsilon_(0)(2a))=sqrt(12)(sigma)/(epsilon_(0))=2sqrt(3)(sigma)/(epsilon_(0))`
`rArr x=3`
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Knowledge Check

  • The densities at points A and B are rho_(0) and (3 rho_(0))/(2) . Find the value of x on P-axis.

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  • An infinite sheet carrying a unifom surface charge density sigma lies on the xy-plane. The work done to carry a charge q from the point vecA=a(hati+2hatj+3hatk) to the point vecB=a(hati-2hatj+6hatk) (where a is a constant with the dimensions of length and epsilon_(0) is the permittivity of free space) is

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