What if a cylindrical capacitor has an inner conductor with a radius of 1.5 mm and an outer conductor with a radius of 3.4 mm. And the two conductors are separated by vacuum, and the entire capacitor is 3.0 m long. a) Then what is the capacitance per unit length? b) If the potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (in magnitude and sign) on both conductors. Find the charge (in magnitude and sign) on both conductors.

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What if a cylindrical capacitor has an inner conductor with a radius of 1.5 mm and an outer conductor with a radius of 3.4 mm. And the two conductors are separated by vacuum, and the entire capacitor is 3.0 m long.  a) Then what is the capacitance per unit length?  b) If the potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (in magnitude and sign) on both conductors. Find the charge (in magnitude and sign) on both conductors.

Talisha · Accepted Answer

a) Inner radius r1=1.5×10−3 m  outer radius r2=3.4×10−3 m  The capacitance of a cylindrical capacitance is given by  C=2πϵ0lln⁡(r2r2)  Cl=2(3.14)(8.854×10−12)ln⁡(3.4×10−3 m1.5×10−3 m)  =6.80×10−11 Fm  (b)  The inner conductor is at a higher potential than the outer conductor it implies that the sign of the charge on the inner conductor is positive outer conductor is negative the magnitude of charge will be same.  The magnitude of charge on each capacitor plate is,  q=(Cl)lvab  =6.80×10−11 FM(3.0m)(350×10−3 V)  =7.14×10−11 C  Charge on the outer conductor q=7.14×10−11 C  Charge on the inner conductor q=−7.14×10−11 C

Jeffrey Jordon · Accepted Answer

Answer:  Part(a): The capacitance per unit length of the capacitor is  1.56×10−10Fm−1  Part(b): The charge in the inner shell is +1.6×10−10C and the charge of the outer shell is −1.6×10−10C  Explanation:  Part(a):  The capacitance (C) per unit length of a cylindrical capacitor is given by  C=2πϵ0log⁡(b/a)  where &#039;ϵ0&#039; is the permittivity of free space, &#039;b&#039; is the radius of the outer shell and &#039;a&#039; is the radius of the inner shell.  Given, b=3.4mm=0.0034 and a=1.5mm=0.0015 mm. We know, ϵ0=8.854×10−12Fm−1. Substituting the values in the above equation,  C=2π×8.85×10−12Fm−1log⁡(0.0034/0.0015)=1.56×10−10Fm−1  Part(b):  As the voltage on the inner conductor is higher than that of the outer conductor, positive charge resides on the inner shell and negative charge resides on the outer shell.  The charge &#039;Q&#039; of a capacitor of length &#039;L&#039; having capacitance &#039;C&#039; and potential difference &#039;V&#039; is given by  Q=CVL  Given, V=350mV=350×10−3V and L=3.0m. Substituting these values in the above expression  Q=1.56×10−10Fm−1×350×10−3V×3m=1.6×10−10C  The charge in the inner shell is +1.6×10−10C and the charge of the outer shell is −1.6×10−10C.

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A cylindrical capacitor has an inner conductor of radius 1.5 mm andan outer conductor of radius 3.4 mm.The two conductors are separated by vacuum, and

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