Gauss\'s Law and the Electric Field A negative charge -Q is uniformly distribute
ID: 1527077 • Letter: G
Question
Gauss's Law and the Electric Field A negative charge -Q is uniformly distributed throughout the spherical volume of +2Q radius R. A positive point charge + 2Q is at the center of the sphere. Determine each of the following in terms of the quantities given and fundamental constants. the electric field E outside the sphere at a distance r > R from the center the electric field inside the sphere at a distance r R). At a point on this Gaussian surface, show the vectors E and dA. Starting from the basic physics principle of Gauss' law, Integral E dA = q_enclosed/epsilon, solve forExplanation / Answer
1) at r > R
iamgine a Gaussian spherical surface with radius r,
charge enclosed by the Gaussian surfcae, Qin = 2*Q - Q
= Q
net elctric flux through the gaussian surface,
integral E.dA = Qin/epsilon
E*A = Q
E*4*pi*r^2 = Q/epsilon
E = Q/(4*pi*epsilon*r^2)
= k*Q/r^2 <<<<<<--------------------Answer
2)
at r < R
iamgine a Gaussian spherical surface with radius r,
volume charge density of sphere, rho = -Q/V
= -Q/( (4/3)*pi*R^3)
charge enclosed by the Gaussian surfcae, Qin = +2*Q + rho*volume of Gaussian sphere
= 2*Q + (-Q/( (4/3)*pi*R^3))*(4/3)*pi*r^3
= 2*Q - Q*r^3/R^3
= Q*(2 - r^3/R^3)
net electric flux through the gaussian surface,
integral E.dA = Qin/epsilon
E*A = Q*(2 - r^3/R^3)
E*4*pi*r^2 = Q*(2 - r^3/R^3)/epsilon
E = Q*(2 - r^3/R^3)/(4*pi*epsilon*r^2)
= k*Q*(2 - r^3/R^3)/r^2
= k*Q*(2/r^2 - r/R^3) <<<<<<--------------------Answer
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