Based on your result to c), compote the potential difference between the What it

Question

Based on your result to c), compote the potential difference between the What it the capacitance of the capacitor? Show work. If the capacitor were filled with a dieleetric of value x=3.0. how would that affect your answers to and c)? (assume Q remains unchanged) A capacitor consisting of conducting coaxial cylinders of radii a=1.0 cm and b=3.0 cm, respectively, and length =25.0 cm is connected to a voltage source, as shown below. When the capacitor is charged, the inner cylinder develops a charge of . Neglect end effects and assume that the region between the cylinders is filled with air (kappa-1.0). Where does the charge on the inner conductor reside? Explain. What charee is on the outer conductor? Explain Use Gauss's Law to determine an expression for the electric field at a distance r from the axis of the cylinder where a

Explanation / Answer

(a) Since both cylinders are conductors , the whole charge resides on the surface of a inner cylinder (b) The same and opposite  charge will induce on the inner surfave of a outer cylinder i.e   - 12 C The charge on the outer surface of a cylinder is + 12C (C) electric field   in the region   ( a < r < b ) According to gauss law charge enclosed by the gaussian surface with radius r E( 2rl ) = Q_enclosed / _0 Here Q_enclosed   Q = + 12 F Electric field E =  Q / 2_0rl (d) Potential difference between the cylinders (a < r < b) V = - E .dr V = - (Q / 2_0l) (1/r) dr   lmit form a to b V = - (Q / 2_0l) ln (b/a )   =  (Q / 2_0l) ln ( b/a   )    V = (12 * 10^-6 F ) /( 2* ( 8.85*10^-12 N.m^2 /C^2 ) * 0.25 m ) ln ( 0.03/0.01 ) V =    9.488*10^5 V capacitence C   = Q / V   = Q / (Q / 2_0l) ln (  b / a    ) )   = 2_0 l  / ln( b/a) C = 12.647*10^-12 F = 12.647 pF (f) capacitence with dielectric C '   = kC    = ( 6.0 )(   12.647 pF) =37.941*pF

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