Physics Assertion (A) & Reason (B) MCQ

3. A is true, but R is false.

Explanation:

If two points P and Q in an electric field are separated by an infinitesimal distance $\triangle\text{x}$ and have a potential difference $\triangle\text{V}$ between then, $\text{E}=\frac{-\triangle\text{V}}{\triangle\text{x}}.$

Here negative sign implies that $\vec{\text{E}}$ has got a direction opposite to the potential gradient i.e. in the direction of $\vec{\text{E}}$ the potential decreases i.e. positive charge always move from a higher potential point to a lower potential point.

2. Both A and R are true, but R is not the correct explanation of A.

Explanation:

When the plates of a capacitor are moved further apart, the capacitance gets decreased. As battery remains connected, hence charge q( = CV) on the plates is decreased, and energy $\text{U}=\Big[\frac{1}{2}\text{CV}^2\Big]$ also decreases. Some charge from the plates flows to the battery i.e. some energy of capacitor is transferred to the battery. Work done against electrostatic attraction between plates is used in the transference of energy and is dissipated in the form of heat energy in connection wires.

1. Both A and R are true, and R is the correct explanation of A.

Explanation:

The dielectric molecules are polarised, producing an opposite electric field. Thus the effective electric field and hence the potential difference between the plates is reduced and consequently, the capacitance is increased $\big(\because\text{C}=\frac{\text{Q}}{\text{V}}\big).$

4. A is false, and R is also false.

Explanation:

On increasing the charge, potential increases. But capacity of a capacitor is fixed by geometry of condenser $\Big(\text{C}=\frac{\in_0\text{A}}{\text{d}}\Big).$ Capacitance is independent of charge.

1. Both A and R are true, and R is the correct explanation of A.

Explanation:

Inside the charged metallic sphere every point is at the same electric potential, hence $\text{W}=\text{q}\triangle\text{V}=0.$ But outside the sphere, there exists a potential gradient at every point, hence $\text{W}\not=0.$

1. Both A and R are true, and R is the correct explanation of A.

Explanation:

The capacity of a parallel plate condenser is given by,

$\text{C}=\frac{\text{q}}{\text{V}}$ (i)

Electric field intensity becomes $\frac{1}{\text{K}}$ times [as $\text{K}=\frac{\text{E}_0}{\text{E}},$] therefore potential V also becomes $\frac{1}{\text{K}}$ times.

Hence, from equation (i) capacity, becomes K times. Thus electric field decreases and capacitance increases when condenser is filled with insulated medium of some dielectric constant.

3. A is true, but R is false.

Explanation:

The capacitance of a capacitor filled partially with a dielectric of thickness t given by,

$\text{C}=\frac{\in_0\text{A}}{\text{d-t}[1-\frac{1}{\text{K}}]}$

For metals, $\text{K}=\infty,$ $\therefore\text{C}=\frac{\in_0\text{A}}{\text{d-t}}.$

Now if the capacitor is filled completely with a metallic slab, then t = d, $\therefore\text{C}=\infty$ i.e., when a charged capacitor filled fully with a metallic slab, then capacitor is short circuited i.e., it will no more work as a capacitor.

1. Both A and R are true, and R is the correct explanation of A.

Explanation:

Capacitance of parallel plate capacitor is

$\text{C}=\frac{\in_0\text{A}}{\text{d}}.$ Thus distance decreases and capacitance of capacitor increases.

3. A is true, but R is false.

Explanation:

The capacitance C, with dielectric between the plates, is given as

$\text{C}=\frac{\in_0\text{KA}}{\text{d}}$

As, $\text{d'}=\frac{\text{d}}{2}\text{K'}=\text{3K},$ then new capacitance becomes

$\text{C'}=\frac{\in_0\text{3KA}}{\frac{\text{d}}{2}}$

$\Rightarrow\text{C'}=\frac{6\in\text{KA}}{\text{d}}=\text{6C}$

The capacitance C depends only on the geometrical configuration (shape, size, separation) of the system of two conductors.