A parallel plate capacitor of capacitance 1 F is charged to a potential difference of 20 V. The distance between plates is 1 F. The energy density between plates of capacitor is : [2025]

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Q.
A parallel plate capacitor of capacitance 1 $μ$ F is charged to a potential difference of 20 V. The distance between plates is 1 $μ$ F. The energy density between plates of capacitor is : [2025]

1 $1.8 \times 10^{3} J / m^{3}$

2 $2 \times 10^{4} J / m^{3}$

3 $2 \times 10^{2} J / m^{3}$

4 $1.8 \times 10^{5} J / m^{3}$

Ans.
(1)

Energy density $u = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} ε_{0} {(\frac{V}{d})}^{2}$

$\Rightarrow u = \frac{1}{2} (8.85 \times 10^{- 12}) {(\frac{20}{10^{- 6}})}^{2} J / m^{3} = 1.8 \times 10^{3} J / m^{3}$

Solution

Q.
A parallel plate capacitor of capacitance 1 $μ$ F is charged to a potential difference of 20 V. The distance between plates is 1 $μ$ F. The energy density between plates of capacitor is : [2025]

1 $1.8 \times 10^{3} J / m^{3}$

2 $2 \times 10^{4} J / m^{3}$

3 $2 \times 10^{2} J / m^{3}$

4 $1.8 \times 10^{5} J / m^{3}$

Ans.
(1)

Energy density $u = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} ε_{0} {(\frac{V}{d})}^{2}$

$\Rightarrow u = \frac{1}{2} (8.85 \times 10^{- 12}) {(\frac{20}{10^{- 6}})}^{2} J / m^{3} = 1.8 \times 10^{3} J / m^{3}$

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Solution

Q. A parallel plate capacitor of capacitance 1 μF is charged to a potential difference of 20 V. The distance between plates is 1 μF. The energy density between plates of capacitor is : [2025]

1 1.8×103J/m3

2 2×104J/m3

3 2×102J/m3

4 1.8×105J/m3

Ans. (1) Energy density u=12ε0E2=12ε0(Vd)2 ⇒ u=12(8.85×10–12)(2010–6)2J/m3=1.8×103J/m3

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Q.
A parallel plate capacitor of capacitance 1 $μ$ F is charged to a potential difference of 20 V. The distance between plates is 1 $μ$ F. The energy density between plates of capacitor is : [2025]

1 $1.8 \times 10^{3} J / m^{3}$

2 $2 \times 10^{4} J / m^{3}$

3 $2 \times 10^{2} J / m^{3}$

4 $1.8 \times 10^{5} J / m^{3}$

Ans.
(1)

Energy density $u = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} ε_{0} {(\frac{V}{d})}^{2}$

$\Rightarrow u = \frac{1}{2} (8.85 \times 10^{- 12}) {(\frac{20}{10^{- 6}})}^{2} J / m^{3} = 1.8 \times 10^{3} J / m^{3}$