Physics M.C.Q (1 Marks)

3. C/ m²

Explanation:

Surface charge density means how much charge is stored in the unit surface area of a conductor.

So, it’s unit will be$=\frac{\text{the unit of the charge}}{\text{the unit of area}}$. In SI the unit of charge is Coulomb and the unit of area is m².

1. M₁ = M_2​

Explanation:

$\theta_1​=θ_2​,$​ if their masses are the same because the force of repulsion due to charges is the same.

The force Mg $\sin\theta$ opposes the force of repulsion. Even if Q₁​ and Q_2​ are different, the force of repulsion is the same for both as they are mutual.

1. The flux of the electric field through the sphere is zero.

3. The electric field is not zero anywhere on the sphere.

Explanation:

$\therefore$ Net charge of electric dipole is always zero.

$\phi=\frac{\text{q}_{\text{inside}}}{\in_0}$ Here inside = 0

The flux of the electric field through the sphere is zero. But the electric field is not zero any where on the sphere. Take 'A' point any where on the sphere, and find the electric field on the surface. You will get non zero electric field on the sphere.

1. Newtons 3rd Law of Motion.

Explanation:

Coulomb's law agrees with Newton's 1st, 2nd and 3rd law of motion. Coulomb's law states that $\text{F}=\text{K}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$ here the two charge particle is similar to the objects of the Newton's 1st law. Newton's 3rd law is same as the two similar charge particles repel to each other. Which coincides with Newton's 3rd law every action there is an equal and opposite reaction.

2. 1 : 1

Explanation:

According to Newton's 3rd law, every action has equal and opposite reaction.

Thus the force exerted by charge on Q due to  2Q is equal but opposite in direction to the force exerted by charge on 2Q due to Q i.e

F_Q​ = F_2Q​

$\Rightarrow\frac{\text{F}_\text{Q}}{\text{F}_\text{2Q}}=1$

4. All the distances

Explanation:

Coulomb's law is true for all distances whether it is small and large. Hence it is called a long range force.

Coulomb's force $\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\in_\circ\text{r}^2}$

$\Rightarrow\text{F}\propto\frac{1}{\text{r}^2}$

4. Is the same for all the figures.

Key concept: According to Gauss' law of electrostatics, the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity,

$\text{i.e.,}\phi=\frac{\text{Q}_\text{enclosed}}{\in_0}$

Thus, electric flux through a surface doesn't depend on the shape, size or area of a surface but it depends on the amount of charge enclosed by the surface.

In given figures the charge enclosed are same that means the electric flux through all the surfaces should be the same. Hence option (d) is correct.

1. 9 × 10³N

Explanation:

Q = 1C r = 1000m

$\text{F}=\frac{\text{KQQ}}{\text{r}^2}=\frac{9\times10^3}{((10)^3)^2}=9\times10^3\text{N}$

1. Total flux through the surface of the sphere is $\frac{-\text{Q}}{\in_0}$.

3. Flux through the surface of sphere due to 5Q is zero.

Gauss' law states that total electric flux of an enclosed surface is given by $\frac{\text{q}}{\in_0}$.

Here, q is the net charge enclosed by the Gaussian surface.

From the figure,

Net charge inside the surface is = Q - 2Q = -Q.

Net flux through the surface of the sphere $=\frac{-\text{Q}}{\in_0}$

Here, charge 5Q lies outside the Gaussian surface, so it will not make no contribution to electric flux through the given surface.

2. $\oint_\text{s}\text{E.dS}=0$ if the charge is outside the surface.

4. $\oint_\text{s}\text{E.dS}=\frac{\text{q}}{\in_0}$ if charges of magnitude q were inside the surface.

According to the Gauss' law

$\int_\text{s}\text{E.dS}=\frac{\text{q}}{\in_0}$

Where q is the charge enclosed by the surface. If the charge is outside the surface, then charge.

If the charge is outside the surface, then charge enclosed by the surface is q = 0, therefore,

$\int_\text{s}\text{E.dS}=0$

Electric flux doesn't depend on the nature or type of charge.

3. $\frac{\text{q}}{2\in_0}$

Explanation:

In the charge 'q' in side closed cylinder, due to charge the flux of the electric field through the surface of the vessel is $\frac{\text{q}}{\in_0}.$ Here A charge q is placed at the centre of the open end of a cylindrical, By symmetricily Half line of flux goes outside & half flux line goes inside. So we can say that the flux of the electric field through the surface of the vessel is $\frac{\text{q}}{2\in_0}.$

1. $4\pi$

Explanation:

A steradian is the solid angle subtended at the center of a sphere of radius r by a section of its surface area of magnitude equal to r². Since the surface area is $4\pi\text{r}^2$, there are $4\pi$ steradians surrounding a point in space.

3. remain same

Explanation:

From Coulombs law

$\text{F}\propto\frac{\text{q}_1\text{q}_2}{\text{r}^2}$

As the charges and distance are same in either cases, the force between the charges remains constant.

4. The torque on the dipole due to the field may be zero.

Explanation:

In the uniform Electric field the net electric force on the dipole is alwas zero.

In uniform Electric field the torque on the dipole due to field may be zero.

T = 0

Here t Þ Torque

$\text{T}=2\text{qEl}\sin\text{q }\otimes\neq0$