A parallel plate air capacitor is connected to a battery. After charging fully, the batter is disconnected and the plates are pulled further apart. The strength of the electric field inside the capacitor would.

remain unchanged

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increase

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decrease

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oscillate

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Solution

The correct option is C decrease
Capacitor has now constant potential.

$E = \frac{E}{d}$ Simple definition

$d ↑\Rightarrow E ↓$

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A parallel plate air capacitor is connected to a battery. After charging fully, the batter is disconnected and the plates are pulled further apart. The strength of the electric field inside the capacitor would.

The correct option is C decreaseCapacitor has now constant potential.E=Ed Simple definitiond↑⇒E↓

The correct option is C decrease
Capacitor has now constant potential.

$E = \frac{E}{d}$ Simple definition

$d ↑\Rightarrow E ↓$