Question

A battery is kept connected to the plates of a parallel-plate capacitor. A dielectric slab of dielectric constant \(K\) is then introduced between the plates such that it covers the entire space between the plates. Choose the correct answer from the given alternatives.

• A. The electric field between the plates will increase.
• B. The charge on the plates will decrease.
• C. The energy stored in the capacitor will decrease.
• D. The electric field between the plates will remain the same.

Hint:

For a capacitor connected to a battery: \[ V=\text{constant}. \] When a dielectric is inserted, \[ C \uparrow,\qquad Q \uparrow,\qquad U \uparrow, \] but \[ E=\frac{V}{d} \] remains unchanged.

Answer 1

The Correct Option is D

Step 1: Spot the constant quantity.
The capacitor stays joined to the battery, so the voltage across the plates is fixed at the battery value $V$ the whole time.

Step 2: Recall the field formula.
For parallel plates with separation $d$, the field between them is \[ E=\frac{V}{d}. \]

Step 3: What the dielectric changes.
Slipping in a slab of dielectric constant $K$ raises the capacitance to $C'=KC$, but it does not change $V$ (the battery holds it) nor the gap $d$.

Step 4: Effect on the field.
Since both $V$ and $d$ are unchanged, \[ E=\frac{V}{d} \] stays exactly the same.

Step 5: Cross-check the other options.
With $V$ fixed, charge $Q=C'V=KCV$ actually rises, and energy $\tfrac12 C'V^2=\tfrac12 KCV^2$ also rises. So the options about field rising, charge falling, or energy falling are all wrong.

Step 6: Final answer.
The field between the plates is unchanged.
\[ \boxed{\text{The electric field between the plates remains the same}} \]