Electrostatic Potential and Capacitance Set-2

Q1. A parallel‑plate capacitor of capacitance $4.0\,\mu\text{F}$ is charged to $200\,\text{V}$ and then disconnected from the battery. A dielectric slab is inserted completely between the plates, increasing the capacitance to $12.0\,\mu\text{F}$. Neglecting fringe effects, the final energy stored in the capacitor is:

Q2. Assertion (A): If the electric potential at a point is $V=0$, then the electric field $\vec{E}$ at that point must be zero.
Reason (R): The electric field is $\vec{E}=-\nabla V$, so $\vec{E}$ depends on the spatial variation of $V$ (its gradient), not on the absolute value of $V$.

Q3. The electric potential along the $x$‑axis is given by

A positive test charge $+1.0\,\mu\text{C}$ is moved slowly from $x=1\ \text{m}$ to $x=5\ \text{m}$. The work done by the electrostatic force on the charge is:

Q4. Two isolated capacitors, $C_1=2.0\,\mu\text{F}$ (initially at $100\ \text{V}$) and $C_2=6.0\,\mu\text{F}$ (initially uncharged), are disconnected from any battery and then connected in parallel (positive to positive) by ideal wires until equilibrium. The energy dissipated as heat during this redistribution is:

Q5. Assertion (A): When a dielectric slab is slowly inserted into a charged isolated (disconnected) parallel‑plate capacitor, the electrostatic energy stored in the capacitor decreases.
Reason (R): The dielectric is attracted into the capacitor by the field so the electrostatic field does positive work on the slab; that work comes from the stored electrostatic energy.

Q6. Two point charges $q_1=+2.0\,\mu\text{C}$ and $q_2=+3.0\,\mu\text{C}$ are brought from infinity and placed $0.10\ \text{m}$ apart in vacuum. The work required to bring them from infinity to this separation (electrostatic potential energy) is:

Q7. A point charge $q=2.0\,\mu\text{C}$ is held at a distance $d=3.0\ \text{cm}$ from an infinite grounded conducting plane. Using the method of images, the magnitude of the attractive force between the charge and the plane is:

Q8. Assertion (A): For a conductor in electrostatic equilibrium the electric potential is the same at every point inside the conductor and on its surface.
Reason (R): In electrostatic equilibrium the electric field inside a conductor is zero, hence $\nabla V=0$ inside the conductor.

Q9. The potential of a spherically symmetric charge distribution is given by

(continuous at $r=0.20\ \text{m}$). The surface charge density $\sigma$ on the spherical surface at $r=0.20\ \text{m}$ is:

Q10. Two capacitors $C_1=3.0\,\mu\text{F}$ and $C_2=6.0\,\mu\text{F}$ are charged separately so that $C_1$ has $+100\ \text{V}$ across it and $C_2$ has $-50\ \text{V}$ across it (i.e., equal and opposite charges in magnitude). Their positive plates are then connected to each other by an ideal wire and allowed to reach equilibrium. The energy dissipated in the process is: