Monday

December 22, 2014

December 22, 2014

Posted by **Anonymous** on Sunday, January 27, 2013 at 12:57pm.

(a) 5.00 cm

- physics -
**Elena**, Sunday, January 27, 2013 at 2:46pmTake the ring of the radius ‘r’ and the width ‘dr’ on the disc.

The electric field at the distance ‘x’ from the center of the disc is

dE=x•dq/4πε₀• {sqrt(r²+x²)}³,

where dq=σ•dA = σ•2•π•r•dr.

E(x)=

=intergral(limits: from 0 to R)

{σ•2•π•r•x•dr/ 4πε₀• [sqrt(r²+x²)]³ =

=(σx/2ε₀)•{(1/x)- [1/sqrt(R²+x²)]}.

- and -
**balls**, Sunday, September 15, 2013 at 2:41pmweiners

- physics -
**66yy6y6**, Thursday, November 20, 2014 at 3:11pmrrtdfrt

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