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A parallel plate capacitor is connected across a 2V battery and charged. The battery is then disconnected and a glass slab is introduced between plates. Which of the following pairs of quantities decreases?
  • a)
    Charge and potential difference
  • b)
    Potential difference and energy stored
  • c)
    Energy stored and capacitance
  • d)
    Capacitance and charge
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A parallel plate capacitor is connected across a 2V battery and charge...
When battery is disconnected, charge remains constant. On introducing glass slab, capacity increases, Potential difference and energy stored decreases.
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Most Upvoted Answer
A parallel plate capacitor is connected across a 2V battery and charge...
Explanation:

When a parallel plate capacitor is charged using a 2V battery, the capacitor stores certain amount of energy. Let Q be the charge on the capacitor, C be the capacitance of the capacitor and V be the potential difference across the plates. The energy stored in the capacitor is given by:

E = 1/2 * QV = 1/2 * CV^2

When a glass slab is introduced between the plates, the capacitance of the capacitor decreases. This is because the presence of a dielectric material between the plates increases the electric field strength, which in turn reduces the capacitance. Let us see how this affects the various quantities:

1. Charge: The charge stored on the capacitor remains the same. This is because the charge on the capacitor is determined by the battery voltage and the capacitance of the capacitor, and is independent of the presence of a dielectric material between the plates.

2. Potential difference: The potential difference across the plates decreases when a dielectric material is introduced between them. This is because the electric field strength reduces due to the presence of the dielectric material, which in turn reduces the potential difference.

3. Energy stored: The energy stored in the capacitor decreases when a dielectric material is introduced between the plates. This is because the capacitance of the capacitor decreases, and the energy stored in a capacitor is proportional to the square of the potential difference and the capacitance.

4. Capacitance: The capacitance of the capacitor decreases when a dielectric material is introduced between the plates. This is because the presence of a dielectric material increases the electric field strength, which in turn reduces the capacitance.

Therefore, the correct answer is option B - Potential difference and energy stored.
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A parallel plate capacitor is connected across a 2V battery and charged. The battery is then disconnected and a glass slab is introduced between plates. Which of the following pairs of quantities decreases?a)Charge and potential differenceb)Potential difference and energy storedc)Energy stored and capacitanced)Capacitance and chargeCorrect answer is option 'B'. Can you explain this answer?
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