Asked by Anonymous
A uniformly charged disk of radius 35.0 cm carries a charge density of 6.70 multiplied by 10-3 C/m2. Calculate the electric field on the axis of the disk at the following distances from the center of the disk.
(a) 5.00 cm
(a) 5.00 cm
Answers
Answered by
Elena
Take 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)]}.
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)]}.
Answered by
balls
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