A medium having dielectric constant K >1 fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is d. The capacitor is connected to a battery of voltage V, as shown in Figure (a). Now, both the plates are moved by a distance of � � from their original positions, as shown in Figure (b). Figure (a) Figure (b) In the process of going from the configuration depicted in Figure (a) to that in Figure (b), which of the following statement(s) is(are) correct? (A) The electric field inside the dielectric material is reduced by a factor of 2𝐾. (B) The capacitance is decreased by a factor of � ��� . (C) The voltage between the capacitor plates is increased by a factor of (𝐾 + 1). (D) The work done in the process DOES NOT depend on the presence of the dielectric material

 We analyze the problem step by step to determine which statements are correct.

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Step 1: Initial configuration (Figure a)

• The capacitor has plates of large area $A$, separation $d$, and is filled with a dielectric material of dielectric constant $K>1$.

• The capacitance $C_1$ is:

  $$C_1=\frac{K{ϵ}_{0}A}{d}\mathrm{.}$$

• The capacitor is connected to a battery of voltage $V$, so the charge $Q$ on the capacitor is:

  $$Q=C_{1}V=\frac{K{ϵ}_{0}A}{d}V\mathrm{.}$$

• The electric field $E_1$ inside the dielectric is:

  $$E_1=\frac{V}{d}\mathrm{.}$$

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Step 2: Final configuration (Figure b)

• The plates are moved by a distance $\frac{d}{2}$ from their original positions, so the new separation between the plates is $d^{\mathrm{\prime }}=d+\frac{d}{2}=\frac{3d}{2}$.

• The capacitance $C_2$ in the new configuration is:

  $$C_2=\frac{K{ϵ}_{0}A}{d^{\mathrm{\prime }}}=\frac{K{ϵ}_{0}A}{\frac{3d}{2}}=\frac{2K{ϵ}_{0}A}{3d}\mathrm{.}$$

• The charge $Q$ remains the same because the capacitor is still connected to the battery. The new voltage $V^{\mathrm{\prime }}$ is:

• The electric field $E_2$ inside the dielectric in the new configuration is:

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Step 3: Analyze the statements

(A) The electric field inside the dielectric material is reduced by a factor of $2K$.

• From Step 2, the electric field $E_2=\frac{V}{d}$, which is the same as $E_1$. Thus, the electric field is not reduced.

• Statement (A) is incorrect.

(B) The capacitance is decreased by a factor of $\frac{K+1}{2K}$.

• The initial capacitance $C_1=\frac{K{ϵ}_{0}A}{d}$, and the final capacitance $C_2=\frac{2K{ϵ}_{0}A}{3d}$.

• The ratio .

• The capacitance is decreased by a factor of $\frac{2}{3}$, not $\frac{K+1}{2K}$.

• Statement (B) is incorrect.

(C) The voltage between the capacitor plates is increased by a factor of $(K+1)$.

• From Step 2, the voltage increases from $V$ to $\frac{3}{2}V$, which is a factor of $\frac{3}{2}$, not $(K+1)$.

• Statement (C) is incorrect.

(D) The work done in the process DOES NOT depend on the presence of the dielectric material.

• The work done in moving the plates depends on the change in energy stored in the capacitor. The energy stored in a capacitor is $U=\frac{1}{2}C{V}^2$.

• The presence of the dielectric material affects the capacitance $C$, and thus the energy stored. Therefore, the work done does depend on the presence of the dielectric material.

• Statement (D) is incorrect.

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Final Answer:

None of the statements (A), (B), (C), or (D) are correct.