The capacity of a parallel plate condenser with air as a dielectric is $2 μ F$ . The space between the plates is filled with a dielectric slab with $K = 5$ . It is charged to a potential of $200 V$ and disconnected from a cell. Work was done in removing the slab from the condenser completely

0.8 J

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0.6 J

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1.2 J

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0.032 J

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Solution

The correct option is C $0.032 J$
Given,

Capacity, $C = 2 μ F = 2 \times 10^{- 6} F$

Voltage, $V = 200 V$

Dielectric constant, $K = 5$

We have,

Work done, $W = \frac{1}{2} C V^{2} (1 - \frac{1}{K})$

$W = \frac{1}{2} \times 2 \times 10^{- 6} \times 200^{2} (1 - \frac{1}{5}) = 0.032 J$

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The capacity of a parallel plate condenser with air as a dielectric is 2μF. The space between the plates is filled with a dielectric slab with K=5. It is charged to a potential of 200V and disconnected from a cell. Work was done in removing the slab from the condenser completely

The correct option is C 0.032JGiven,Capacity, C=2μF=2×10−6FVoltage, V=200VDielectric constant, K=5We have,Work done, W=12CV2(1−1K)W=12×2×10−6×2002(1−15)=0.032J

The capacity of a parallel plate condenser with air as a dielectric is $2 μ F$ . The space between the plates is filled with a dielectric slab with $K = 5$ . It is charged to a potential of $200 V$ and disconnected from a cell. Work was done in removing the slab from the condenser completely

The correct option is C $0.032 J$
Given,

Capacity, $C = 2 μ F = 2 \times 10^{- 6} F$

Voltage, $V = 200 V$

Dielectric constant, $K = 5$

We have,

Work done, $W = \frac{1}{2} C V^{2} (1 - \frac{1}{K})$

$W = \frac{1}{2} \times 2 \times 10^{- 6} \times 200^{2} (1 - \frac{1}{5}) = 0.032 J$