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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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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 the entire capacitor is 3.0 m long.
a) What is the capacitance per unit length?
b)The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors. Find the charge (magnitude and sign) on both conductors.

2021-03-06
a) Inner radius $$\displaystyle{r}_{{1}}={1.5}\times{10}^{{-{3}}}$$ m
outer radius $$\displaystyle{r}_{{2}}={3.4}\times{10}^{{-{3}}}$$ m
The capacitance of a cylindrical capacitance is given by
$$\displaystyle{C}={\frac{{{2}\pi\epsilon_{{0}}{l}}}{{{\ln{{\left({\frac{{{r}_{{2}}}}{{{r}_{{2}}}}}\right)}}}}}}$$
$$\displaystyle{\frac{{{C}}}{{{l}}}}={\frac{{{2}{\left({3.14}\right)}{\left({8.854}\times{10}^{{-{12}}}\right)}}}{{{\ln{{\left({\frac{{{3.4}\times{10}^{{-{3}}}\ {m}}}{{{1.5}\times{10}^{{-{3}}}\ {m}}}}\right)}}}}}}$$
$$\displaystyle={6.80}\times{10}^{{-{11}}}\ {\frac{{{F}}}{{{m}}}}$$
(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,
$$\displaystyle{q}={\left({\frac{{{C}}}{{{l}}}}\right)}{l}{v}_{{{a}{b}}}$$
$$\displaystyle={6.80}\times{10}^{{-{11}}}\ {\frac{{{F}}}{{{M}}}}{\left({3.0}{m}\right)}{\left({350}\times{10}^{{-{3}}}\ {V}\right)}$$
$$\displaystyle={7.14}\times{10}^{{-{11}}}\ {C}$$
Charge on the outer conductor $$\displaystyle{q}={7.14}\times{10}^{{-{11}}}\ {C}$$
Charge on the inner conductor $$\displaystyle{q}=-{7.14}\times{10}^{{-{11}}}\ {C}$$