. The chocolate crumb mystery, continued. (a) From your answer to part (a) of th

Question

. The chocolate crumb mystery, continued. (a) From your answer to part (a) of the previous problem* find an expression for the electric potential as a function of the radial distance r from the center of the pipe. (The electric potential is zero on the grounded pipe’s outside wall.) (b) For the typical volume charge density = -1.1 x 10-3 C/m3 , what is the difference in the electric potential between the pipe’s center and its inside wall? Remember R = 5.0 cm.

(answer to chocolate crumb part 1) (a) E (r)= (r)/(2E0 ), inward

Explanation / Answer

electric field and potnetial V are related by E = V/r

here Electric potnetial V = integral E.dr   from r to R

so 0 - V1 = integration from r to R   rho r/2 eo

V1 = rho /2eo *( r^2/2)   from r to R

V1 = rho /4eo * ( R^2 - r^2) ------------<<<<<<<<<<<<<<<Answer

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part C:

potential Difference = Vcentre - V

DV = rho * ( R^2 -0)/4 e0

DV = rho R^2/4 eo

DV = -1.1e -3 * 0.05* 0.05/(4* 8.85 e -12)

DV = potential difference = - 7.76 *10^4 Volts   -----<<<<<<<<<<<<<Answer

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