CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 4 - MCQExams.com

For equilateral triangle AOB, $$AO=OB=BA=a$$  and $$ AC=OC=BC=r$$

The potential at centroid C is $$V=k\dfrac{2q}{AC}+k\dfrac{-q}{OC}+k\dfrac{-q}{BC}=k\dfrac{2q}{r}-k\dfrac{q}{r}-k\dfrac{q}{r}=0$$

The net electric field  at centroid C is $$E=k\dfrac{2q}{r^2}-k\dfrac{q}{r^2}\cos30-k\dfrac{q}{r^2}\cos30=k(0.2)\dfrac{2q}{r^2} \neq 0$$

As the total charge is zero so for dipole moment the choice of origin is independent. We assume O as origin.

dipole moment , $$p=-q(0)-q(L\hat i)+2q(\dfrac{L}{2}\hat i+\dfrac{\sqrt{3}L}{2}\hat j)$$

or $$p=\sqrt{3}qL \hat j$$

Electric field due to charge sheet for charge density $$+\sigma$$ is $$E=\dfrac{+\sigma}{2\epsilon_0}$$ and for $$-\sigma$$ is $$E=\dfrac{-\sigma}{2\epsilon_0}$$

outside the plates field is $$E=\dfrac{+\sigma}{2\epsilon_0}+\dfrac{-\sigma}{2\epsilon_0}=0$$

between the plates field is $$E=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}=\dfrac{\sigma}{\epsilon_0}$$

Ans: (A),(C)

Hint: Find the net force on the dipole by using $$F=qE$$ to calculate the force on both charges of the dipole.
Explanation:
Step 1: Reasons for incorrect options:

• For net electric force to be zero on the dipole, the electric field lines should be parallel to each other, and the dipole is placed in such a way that it is perpendicular to the electric field. But parallel field lines can be there as well for a point charge. So, options A and B are incorrect.
• Now, torque on the dipole is given by the formula:

$$\vec{ \tau} = \vec{P} \times \vec{E}$$

 If the dipole moment $$P$$ and electric field line $$E$$ is in the same direction, in that case, torque will be zero in another case, it will have some non-zero value. So, option C is incorrect.

Answer:

Hence, option D is the correct answer. 

Consider a Gaussian cynlinder of radius $$r$$ and length $$l$$. Using Gauss's law the electric field at distance $$r$$ is $$\displaystyle E.(2\pi rl) =\dfrac{\lambda l}{\epsilon_0} \Rightarrow E=\dfrac{\lambda}{2\pi\epsilon_0 r}$$

Here the electrostatic force is equal to centrifugal force. i.e, 

$$\displaystyle qE=\dfrac{mv^2}{r}$$

$$\displaystyle q\dfrac{\lambda}{2\pi \epsilon_0 r}=\dfrac{mv^2}{r}$$ 

or$$\displaystyle v^2=\dfrac{q\lambda}{2\pi\epsilon_0 m}$$

so $$v$$ is independent of $$r$$.

The electric field at a distance r from the long wire is

$$\displaystyle E=\dfrac{1}{4\pi \epsilon_0} \dfrac{2\lambda}{r}=9\times 10^9\times \dfrac{2\times 2.5\times 10^{-6}}{5\times 10^{-2}}=9\times 10^5 NC^{-1}$$

Hint: Find the net force on the dipole by using $$F=qE$$ to calculate the force on both charges of the dipole.
Explanation:
Step 1: Concept used:
Dipole consists of 2 equal and opposite charges $$+q \ and -q$$.

Consider a dipole, as shown below in the figure.

In uniform electric field E, if we place any dipole at that time,

Step 2: Calculation of force:

Force on $$+q $$ is $$q E$$ and force on charge $$-q$$ is $$-q E$$.

So, the total force will be sum of these two forces, and it will result in zero.

Answer:

Hence, option A is the correct answer. 

Consider a surface of a cylinder of radius $$r$$ and length $$l$$ as Gaussian surface. Apply Gauss's law, $$\displaystyle \vec{E}.\vec{ds}=\dfrac{Q_{en}}{\epsilon_0}$$  where $$Q_{en}$$ is the charge enclosed by the Gaussian surface.  here $$Q_{en}=\lambda l$$

thus, $$\displaystyle E.(2\pi rl)=\dfrac{\lambda l}{\epsilon_0}$$

$$\displaystyle \therefore E=\dfrac{\lambda }{2\pi \epsilon_0 r}=k\dfrac{2\lambda }{r}$$  where $$\displaystyle k=\dfrac{1 }{4\pi \epsilon_0 r}$$

The electric field due to the plate is $$E=\dfrac{\sigma}{2\epsilon_0}$$

Here the force between the charged particle of charge q and the plate is

 $$F=qE=\dfrac{q\sigma}{2\epsilon_0}=\dfrac{2\times 10^{-6}\times 4\times 10^{-6}}{2\times 8.854\times 10^{-12}}=0.45  N$$

Using Gauss's law, the electric flux through the cube is $$\displaystyle \phi=\dfrac{Q_{en}}{\epsilon_0}$$  where $$Q_{en}$$ is the charge enclosed by cube.

As cube has six faces so the flux through each face is $$\displaystyle \phi_e=\dfrac{\phi}{6}=\dfrac{Q_{en}}{6\epsilon_0}$$.

The flux is also the number of lines of force passing through the surface of cube. So 

the number of lines of force through each face of the cube is $$\displaystyle n=\phi_e=\dfrac{Q_{en}}{6\epsilon_0}=\dfrac{0.1\times 10^{-6}}{6\times 8.85 \times 10^{-12}} \sim 1883$$

Hint: Like charges repel each other and unlike charges attract each other.

Part 1 : Figure for given situation

Consider that the soap is in negative charge, 

$$\textbf{Part2: Explanation}$$

$$\bullet$$ The charge will be distributed uniformly over the surface of the bubble by symmetry.

$$\bullet$$ As we know, they want to step much farther apart, as charges repel, the only path outward, taking with them the soap surface.

$$\bullet$$ At the same point, the rise in the restored force of the soap film (surface tension) would be equal and contrary to the electrostatic force, which will result in a new (larger) radius of equilibrium.

$$\bullet$$ Because of the ionic similarities, this can happen to both positive and negatively charged bubbles.Therefore, when a negative charge is given to the soap bubble, then its radius will increase.

Hence option $$A$$ correct.

The potential at a distance r due to dipole is $$\displaystyle V=\dfrac{1}{4\pi\epsilon_0}  \dfrac{\vec{p}.\hat{r}}{r^2}$$  where $$\vec{p}=$$ dipole moment.

or $$\displaystyle V=\dfrac{1}{4\pi\epsilon_0}  \dfrac{p\cos\theta}{r^2}$$ 

or $$\displaystyle p=\dfrac{4\pi\epsilon_0 Vr^2}{\cos\theta}=\dfrac{4\pi\times 8.85\times 10^{-12}\times 2\times 10^{-5}\times (0.1)^2}{\cos30}=2.57 \times 10^{-17}  Cm$$