A variable capacitor consists of two thin coaxial metal cylinders of radii a and b, with...
A variable capacitor consists of two thin coaxial metal cylinders of radii a and b, with (b - a) << a, free to move with respect to each other in the axial direction. The length of the cylinders is L, and the potential difference between the two cylinders is V. Initially, the inner cylinder (radius = a) is completely enclosed by the outer cylinder (radius = b). Using energy methods, find the magnitude and direction of the force on the inner cylinder when it is displaced along the axial direction by a small amount with respect to the outer one. Explain how this force arises.
Homework Answers
A variable capacitor of two thi coaxial metal cylinder of rarii a and b, length L
the capacitor has a line charge lamda
The potential between two capacitor V = (lamda /eps0 ) ( ln b/a)
Work done on the capacitor W = - 1/2 CV2 = -1/2
Q2/C
C = Q/V = lamda L / ( lamda / eps0) (ln b/a)
C = 2 pi eps0 L / (ln/b/a)
When inner cylinder displaced by a distance y along axis, the capacitance will changes with potential
V = ( lamda L /2pi eps0 y ) ln b + (lamda L / 2pi eps0 (L -y) ) ( ln b/a)
capacitance C = Q/ V
= lamda L / ( lamda / 2pi eps0 y) ln b ) + Lamda L / ( lamda /( 2 pi eps0 (L-y) ) ln (b/a) )
C = 2pi eps0 L y / ( ln b ) + 2pi eps0 L (L -y) / ln b/a
Work done in capacitor W = 1/2 Q2/C
F = - dW/ dy
= - dW/ dC x dC/dy
= - 1/2 Q2 ( d/dC (1/C) x dC/dy
F = 1/2 Q2/C2 x dC/dy
F = 1/2 V2 dC/dy
sinc , dC/dy = 2pi eps0 L / lnb + 2pi eps0 L / ln b/a
F = 1/2 V2 ( ( 2pi eps0 L / ln b ) + 2pi eps0 L / ln b/a )
F = pi eps0 V2 L ( 1/ln b + 1/ ln b/a )
The force is an attractive. In outer cylinder line charge produce +Sigma in outer surface and -Sigma in inner surface. It induce the +Sigma in outer surface of inner cylinder and - Sigma in inner surface. this will produce opposite electric field and attract each other.
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