Wednesday

October 22, 2014

October 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

**Answer this Question**

**Related Questions**

Physics - A charged disk of radius R that carries a surface charge density σ...

PHYSICS - A charged disk of radius R that carries a surface charge density ó ...

PHYSICS - Consider a disk of radius 10 cm and positive surface charge density +3...

Physics - Consider a disk of radius 10 cm and positive surface charge density +3...

Physics - A thin insulating thread of length L is attached to a point on the ...

Physics - Consider a disk of radius 2.6 cm with a uniformly distributed charge ...

physics - A compact disk player has just been turned off. A disk decelerates ...

Physics Help - Charge is distributed uniformly along the x axis with density ßx ...

physics - If the magnitude of the electric field in air exceeds roughly 3 106 N/...

Physics - A flat uniform circular disk (radius = 2.10 m, mass = 1.00 102 kg) is ...