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Chapter 11: Problem 1624

A variable condenser is permanently connected to a \(100 \mathrm{~V}\) battery. If capacitor is changed from \(2 \mu \mathrm{F}\) to \(10 \mu \mathrm{F}\). then energy changes is equal to (A) \(2 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(6.5 \times 10^{-2} \mathrm{~J}\) (D) \(4 \times 10^{-2} \mathrm{~J}\)

Short Answer

Expert verified
The change in energy when the capacitor is changed from $2 \mu \mathrm{F}$ to $10 \mu \mathrm{F}$ is \(\boldsymbol{4 \times 10^{-2} \mathrm{~J}}\).

Step by step solution

01

Calculate the energy for the first capacitor

The energy stored in a capacitor can be calculated using the following formula: \[E = \frac{1}{2}CV^2\] where C is the capacitance, and V is the voltage across the capacitor. For the first capacitor with a capacitance of 2 μF, the energy stored can be calculated as: \(E_1 = \frac{1}{2}(2\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\)
02

Calculate the energy for the second capacitor

Similarly, for the second capacitor with a capacitance of 10 μF, the energy stored can be calculated using the same formula: \(E_2 = \frac{1}{2}(10\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\)
03

Calculate the change in energy

Now, we will calculate the change in energy by finding the difference between the two energy values: ΔE = E₂ - E₁
04

Substitute the values and find the answer

Substitute the values of E₁ and E₂ in the formula, and calculate the change in energy: ΔE = \(\frac{1}{2}(10\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\) - \(\frac{1}{2}(2\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\) ΔE = \(4\times10^{-2}\,\mathrm{J}\) Hence, the change in energy is \(\boldsymbol{4 \times 10^{-2} \mathrm{~J}}\), which corresponds to option (D).

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Most popular questions from this chapter

For the system shown in figure, if the resultant force on q is zero, then \(q=\ldots \ldots \ldots\) (A) \(-2 \sqrt{2} \mathrm{Q}\) (B) \(2 \sqrt{2} \mathrm{Q}\) (C) \(2 \sqrt{3} \mathrm{Q}\) (D) \(-3 \sqrt{2} Q\)

Two metal plate form a parallel plate capacitor. The distance between the plates is \(\mathrm{d}\). A metal sheet of thickness \(\mathrm{d} / 2\) and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (A) \(4: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) \(5: 1\)

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The displacement of a charge \(Q\) in the electric field $E^{-}=e_{1} i \wedge+e_{2} j \wedge+e_{3} k \wedge\( is \)r^{-}=a i \wedge+b j \wedge$ The work done is \(\ldots \ldots\) (A) \(Q\left(e_{1}+e_{2}\right) \sqrt{\left(a^{2}+b^{2}\right)}\) (B) \(Q\left[\sqrt{ \left.\left(e_{1}^{2}+e_{2}^{2}\right)\right](a+b)}\right.\) (C) \(Q\left(a e_{1}+b e_{2}\right)\) (D) \(\left.Q \sqrt{[}\left(a e_{1}\right)^{2}+\left(b e_{2}\right)^{2}\right]\)
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