Physics Assertion (A) & Reason (B) MCQ

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

Explanation:

The dielectric constant of any material is $\text{K}=\frac{\text{E}_0}{\text{E}}=\frac{\frac{\text{F}_0}{\text{q}}}{\frac{\text{F}}{\text{q}}}=\frac{\text{F}_0}{\text{F}}$

$\Rightarrow\text{F}=\frac{\text{F}_0}{\text{K}},$ where F₀ is force when conductor is not present between the charge. Fis a force after introduction of conductor between charges. Since dielectric constant of a conducting medium is infinity therefore F = 0.

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

Explanation:

The electric field intensity is equal to force experienced by unit positive test charge q₀ placed at that point i.e.

$\vec{\text{E}}=\frac{\vec{\text{F}}}{\text{q}_0}$ thus $\vec{\text{E}}$ is also a vector quality,

As, $\text{E}=\frac{\text{F}}{\text{q}}=\frac{\text{newton}}{\text{coulomb}}.$

3. A is true but R is false.

Explanation:

According to Coulomb's law, $\text{F}=\text{k}\frac{\text{Q}_1\text{Q}_2}{\text{r}^2}.$

The force on q due to A,

$\text{F}_\text{A}=\frac{1}{4\pi\in_0}\cdot\frac{\text{Qq}}{(\frac{\text{r}}{2})^2}$ to the right

Due to B,

$\text{F}_\text{B}=\frac{1}{4\pi\in_0}\cdot\frac{\text{Qq}}{(\frac{\text{r}}{2})^2}$ to the right

$\therefore$ Their sum is zero whether q is $+\frac{\text{Q}}{4}$or $-\frac{\text{Q}}{4}$ or any other value.

Therefore, it is not true that the third charge has to be $\frac{\text{Q}}{4}$ only. It can be any value.

4. A is false and R is also false.

Explanation:

Surface of a charged conductor is always an equipotential surface, whatever may be its shape. Hence, $\sigma\text{R}=$ constant, at every point on the surface of charged conductor i.e. at the sharpest point (R → 0) of the surface, charge density will be maximum. A uniformly charged conductor exerts no electrostatic force on a point charge located anywhere inside the conductor or electric field is zero.

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

Explanation:

As F = qE, and charge (q) on an electron and on a proton has the same magnitude, therefore force (F) on each is same. Now, acceleration $\text{a}=\frac{\text{F}}{\text{m}}.$

Mass of electron $=\frac{1}{1836}$ mass of a proton. Therefore, acceleration of electron is 1836 times that of a proton.

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

Explanation:

Work done will be zero because in rotating the charge in a circle, force is along the radius, and direction of motion is perpendicular to it.

$\therefore$ Work done, $\text{W}=\vec{\text{F}}\cdot\vec{\text{S}}=\text{FS}\cos\theta=\text{FS}\cos90^\circ=0$

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

Explanation:

According to Gauss's theorem in electrostatics,$\phi=\frac{\text{q}}{\in_0}$

$\phi=\frac{\text{q}}{\in_0}=8.85\times10^{-12}[8\times10^3-2\times10^3]$

$=53.10\times10^{-9}\text{C}=0.053\mu\text{C}.$

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

Explanation:

The electric flux through the cube, $\phi=\frac{\text{q}}{\in_0}.$ A cube has six faces of equal area. Therefore, electric flux through each face $=\frac{1}{6}\phi=\frac{1}{6}(\frac{\text{q}}{\in_0}).$

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

Explanation:

Coulomb's force or electric force given by Coulomb's law is valid over nuclear distances $(\cong10^{-15}\text{m})$ as well as over very large distances.

From Coulomb's law, $\text{F}=\frac{\text{k }\mid\text{ q}_1 \parallel\text{ q}_2 \mid}{\text{r}^2}.$ This generally holds only for charged objects whose size are much smaller than the distance between them.

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

Explanation:

As, $\sigma_1=\sigma_2$ (Given)

$\therefore\frac{\text{q}_1}{4\pi\text{r}_1^2}=\frac{\text{q}_2}{4\pi\text{r}_2^2},$

Or $\frac{\text{q}_1}{\text{q}_2}=\frac{\text{r}_1^2}{\text{r}^2_2}$

[Let r₁ and r₂ be two different radii) Then the ratio of electric field intensities near the surface of spherical conductors,

$\frac{\text{E}_1}{\text{E}_2}=\frac{\text{q}_1}{4\pi\in_0\text{r}^2_1}\times\frac{4\pi\in_0\text{r}^2_2}{\text{q}_2}$

$=\frac{\text{q}_1}{\text{q}_2}\times\frac{\text{r}^2_2}{\text{r}^2_1}=\frac{\text{q}_1}{\text{q}_2}\times\frac{\text{q}_2}{\text{q}_1}=1$

i.e. E₁ = E₂.