Chapter 4: Q30E (page 810)

A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting tube with an inner radius of 2.00 mm. The two conductors are separated by air and charged to a potential difference of 6.00 V. Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

Expert verified

(a) When two conductors separated by air is charged to a potential difference of , then the length for the capacitor is $6.53 \times 10^{- 10} C / m$

(b)The total charge of on the capacitor is $2.3 \times 10^{- 10} C$

(d) The energy stored in the capacitor when fully charged is $6.9 \times 10^{- 10} J$

Step by step solution

Given,

$L = 0.305 m r_{a} = 1.20 m r_{b} = 2.0 m V = 6.0 V$ Now, we know that the capacitor of the cylindrical capacitor is:

$C = \frac{Q}{V} = \frac{2 {πε}_{0} L}{I n (\frac{r_{b}}{r_{a}})}$

Switching V and L we get:

$λ = \frac{Q}{L} = \frac{2 {πε}_{0} V}{I n (\frac{r_{b}}{r_{a}})}$

Plugging in the values we get:

$λ = \frac{2 π \times 8.85 \times 10^{- 12} \times 6}{I n (\frac{2}{1.2})} = 6.52 \times 10^{- 10} C / m$

Therefore, the total charge on the cylindrical capacitor is $6.52 \times 10^{- 10} C / m$

The total charge contained on the cylinder is given by:

$Q = λ L = 6.53 \times 10^{- 10} \times 0.350 = 2.3 \times 10^{- 10} C$

Now, the capacitance is given by:

$C = \frac{Q}{V} = \frac{2.3 \times 10^{- 10}}{6} = 3.83 \times 10^{- 11} F$

Therefore, the capacitance is $3.83 \times 10^{- 11} F$

Now, the energy stored is given by:

$U = \frac{1}{2} Q V = \frac{1}{2} (2.3 \times 10^{- 10}) \times 6 = 6.9 \times 10^{- 10} J$

Therefore, the energy stored is $6.9 \times 10^{- 10} J$

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