Electrostatic Potential and Capacitance Set-1

Q1. A parallel-plate capacitor (plate area $A$, separation $d$) is connected to an ideal battery that maintains a constant potential difference $V$. A dielectric slab of relative permittivity $k>1$ is inserted so that it fills exactly half the gap (thickness $d/2$) while covering the entire plate area; fringing is negligible. After equilibrium is reached, which of the following statements is correct?
I. The magnitude of electric field in the dielectric is $E_d=\dfrac{\sigma}{k\varepsilon_0}$ and in the air gap $E_a=\dfrac{\sigma}{\varepsilon_0}$, where $\sigma$ is the free surface charge density on the plates.
II. The electric displacement $D$ is the same in both layers and equals $\sigma$.
III. The potential drop across the dielectric is smaller than that across the air gap.

Q2. Assertion (A): Two identical capacitors, one initially charged to potential $V$ and the other initially uncharged, are connected in parallel (positive to positive) and isolated; the total electrostatic energy stored after connection is half the initial energy stored in the charged capacitor.
Reason (R): The 'missing' half of the energy is always dissipated as heat in the connecting wire during the redistribution of charge.

Q3. A parallel-plate capacitor with plate area $A=1.0\times10^{-2}\,\mathrm{m^2}$ and plate separation $d=1.0\times10^{-3}\,\mathrm{m}$ is connected across a potential difference $V=100\,$V. Neglect fringing. Using $\varepsilon_0=8.854\times10^{-12}\,\mathrm{F\,m^{-1}}$, the magnitude of surface charge density $\sigma$ on the positive plate is approximately:

Q4. Two isolated capacitors $C_1=3.0\,\mu\mathrm{F}$ and $C_2=6.0\,\mu\mathrm{F}$ are prepared so that $C_1$ is charged to $120\,$V while $C_2$ is uncharged. They are then connected in parallel (positive to positive) and isolated from any battery. The energy dissipated during charge redistribution is most nearly:

Q5. Assertion (A): Two isolated conducting spheres of radii $R_1$ and $R_2$ are connected by a thin conducting wire; at electrostatic equilibrium the charges $Q_1$ and $Q_2$ on the spheres satisfy $\dfrac{Q_1}{Q_2}=\dfrac{R_1}{R_2}$.
Reason (R): During charge redistribution charges flow until the surface charge densities on the two spheres become equal.

Q6. Two radial plots of electrostatic potential $V$ versus distance $r$ from the centre are sketched for charged spherical objects of the same radius $R$ (qualitative shapes described below).
Curve P: $V(r)$ is constant for $0\le r\le R$ and for $r>R$ falls as $V\propto 1/r$.
Curve Q: $V(r)$ is finite at $r=0$, decreases smoothly (no flat region) for $0\le r\le R$, and for $r>R$ also falls as $V\propto 1/r$.
Which identification is correct?

Q7. A parallel-plate capacitor with capacitance $C$ is charged to potential $V$ and then disconnected from the battery so that charge $Q$ is fixed. The plate separation is then reduced quasistatically from $d$ to $d/2$ (plates remain parallel). The final electrostatic energy stored $U_f$ compared to the initial $U_i$ is:

Q8. The graph shows two possible dependences of stored energy $U$ on plate separation $d$ (plate area fixed): Curve X is linear, $U\propto d$, while Curve Y follows $U\propto 1/d$. Which of the following correctly identifies the physical condition producing each curve for a parallel-plate capacitor?

Q9. Assertion (A): For an isolated conducting shell of spherical shape that contains a single point charge $q$ placed off‑centre inside an internal cavity, the electric field outside the conductor is identical to that produced by a point charge $q$ placed at the geometric centre of the conductor.
Reason (R): The uniqueness theorem implies the external potential is determined only by the total enclosed charge and the conductor being an equipotential forces the induced charge on the outer surface to arrange so the external field mimics a central point charge.

Q10. A parallel-plate capacitor has plate area $A=4.0\times10^{-3}\,\mathrm{m^2}$, plate separation $d=1.0\times10^{-3}\,\mathrm{m}$ and plate length along insertion direction $L=0.10\,\mathrm{m}$. It is connected to a battery that maintains $V=200\,$V. A dielectric slab of relative permittivity $k=5$ and thickness equal to $d$ is partially inserted between the plates; for insertion depth $x$ the capacitance is $C(x)=C_0\!\left(1+\dfrac{(k-1)x}{L}\right)$ where $C_0=\dfrac{\varepsilon_0 A}{d}$. The insertion force (towards increasing $x$) is $F=\dfrac{1}{2}V^2\dfrac{dC}{dx}$ and is independent of $x$. The magnitude of $F$ is closest to: