Electric Field

To find the electric field at the center of a cube due to point charges placed at its vertices, we can use symmetry arguments and the principle of superposition.

1. Description of the System:

A cube has a side length of with charges of at each vertex, except for the origin, which has a charge of . The center of the cube is located at .

2. Vector Position of Center from Each Vertex:

• The displacement vector from any vertex with charge to the center is of the form .
• For the charge at the origin, the displacement vector to the center is .

3. Electric Field Due to a Point Charge:

The electric field due to a point charge at a distance is given by:

Here, is the unit vector in the direction of the point from the charge.

4. Evaluate Electric Field at Cube Center:

• The contributions from charges at symmetric positions cancel due to symmetry.
• The only net contribution is from the charges at the origin and the opposite vertex.

Considering the symmetry and calculating the vector sum, the resultant electric field due to charges at all vertices, considering the symmetry and superposition, results in:

5. Logical Conclusion:

Therefore, the electric field at the center of the cube is , which matches the given correct answer.

Electric field at P will be E=

So,

So, d = 3 m

The correct option is (B): 3.0 m

• Potential barrier,
• Width of depletion region,

The relationship between the charges and radii is given by:

Simplifying, we find:

Using the charge conservation equation:

Substitute :

Substitute :

The surface charge densities are related by:

Final Answer:

The correct option is (B): 2

I₂=I4

I₃=0

and I₂+I₄+I₃=I₁

The correct option is (C): Drifting from lower potential to higher potential

the electron starts drifting from the lower potential to higher potential.

drift velocity =

In a metallic conductor, free electrons are always in random motion due to thermal energy. When an electric field is applied, the electrons experience a force in the direction opposite to the field (since electrons are negatively charged).
The velocity of the electrons may make any angle with the acceleration vector between successive collisions. Due to this, the electron's motion is not in a straight line but rather along curved paths. This curved path results from the fact that the electron is continuously undergoing collisions with atoms in the conductor, which changes its direction and velocity.
The net motion of the electrons is not uniform, but they drift slowly towards the positive end (higher potential) of the conductor due to the applied electric field. The term "drift velocity" refers to the average velocity of electrons moving towards the positive terminal under the influence of the electric field.
Thus, the electrons follow curved paths as they move from lower potential to higher potential. This is because their velocity and acceleration vectors are not aligned along the same direction due to the random collisions between electrons and atoms in the conductor.
Step 1: Random motion of electrons:
In the absence of an electric field, electrons in a conductor move randomly in all directions due to thermal motion.
Step 2: Effect of the electric field:
When an external electric field is applied, it exerts a force on the electrons, causing them to drift in the direction opposite to the field. However, since electrons are subject to frequent collisions, their velocity is always changing, and they follow curved paths.
Step 3: Drift of electrons:
The resultant motion of the electrons, due to the superposition of their random motion and drift motion caused by the electric field, results in an overall drift from lower potential to higher potential along curved paths.
Conclusion:
Therefore, the free electrons move in curved paths from lower potential to higher potential.

Step 1: Formula for work done - Work done is given by: where . - Initial potential energy: - Final potential energy:

Step 2: Substitute the values - Substituting values: Simplifying:

Final Answer: The work done is 18 mJ.

Step 1: Analyze the electric field inside and outside the sphere.
- Inside the sphere : Electric field increases linearly with r due to the uniformly distributed charge.
- Outside the sphere : Electric field decreases as , behaving like a point charge at the center.

Step 2: Draw the graph.
- Combine the two observations to get the correct graph for the electric field variation.

Final Answer: The correct graph is represented by option (4).

The torque acting on an electric dipole in a uniform electric field is given by:

where:

• is the dipole moment,
• is the electric field intensity,
• is the angle between and .

Step 1: Calculate the dipole moment ( ).
The dipole moment is:

Step 2: Substitute values into the torque formula.
The electric field is given in dyne/C. Convert it to SI units: Now substitute the values:

Step 3: Simplify the expression.
so:

Solution:
Using Gauss's Law: Since the electric dipole has no net charge (it consists of equal and opposite charges), the net charge . Hence, the net flux . Thus, Assertion A is true. Reason R states that an electric dipole consists of two equal and opposite charges, which is true. Thus, both Assertion A and Reason R are true, and Reason R correctly explains Assertion A. Therefore, the correct answer is .

Given:

The potential due to a point charge is given by the formula: where is the Coulomb constant, is the charge, and is the distance.

Step 1: Potential due to charge

The potential due to the charge is given by:

Step 2: Potential due to

The potential due to charge is calculated as:

Step 3: Using (i) and (ii)

The net potential is obtained by adding the potentials due to and :

Step 4: Simplification and calculation

Step 5: Finding

Given , we conclude:

Final Answer:

To determine the electric field at point 'O' due to the arc 'ABC', we need to analyze the contribution of each segment of the arc to the net electric field.

1. Electric Field Due to Arc 'AB':
   • Given that the electric field at 'O' due to arc 'AB' is .
   • This electric field vector points along the bisector of the angle subtended by the arc 'AB' at the center 'O'.
2. Electric Field Due to Arc 'BC':
   • The arc 'BC' is symmetric to 'AB' about the line 'OC'.
   • Hence, the electric field due to arc 'BC' at 'O' also has magnitude and points along the bisector of the angle subtended by arc 'BC' at 'O'.
3. Net Electric Field at 'O':
   • The electric fields due to arcs 'AB' and 'BC' are perpendicular to each other because 'AB' and 'BC' are separated by a right angle.
   • Therefore, the resultant electric field is the vector sum of these perpendicular electric fields.
   • The magnitude of the resultant electric field is given by:

Thus, the magnitude of the electric field at 'O' due to the arc 'ABC' is .

We have a uniformly charged ring of radius lying in the -plane (center at origin). We seek the point on the positive -axis where the axial electric field is maximum.

Concept Used:

The magnitude of the electric field on the axis of a uniformly charged ring is

directed along for . To maximize with respect to , we differentiate and set the derivative to zero.

Step-by-Step Solution:

Step 1: Maximize the function (the constant factor does not affect the location of the maximum).

Step 2: Solve for :

Final Computation & Result

With ,

The electric field is maximum at (along the positive -axis).

We are given the potential difference between two plates as , and the separation between the plates is 10 cm. The distance between points A and B is 3 cm and 4 cm, respectively, with the total distance between the plates being 10 cm. Using , where is the electric field and is the distance: From the diagram, we know that . 
The potential difference between points A and B is: Thus, the potential difference between points A and B is .

In considering the given Assertion (A) and Reason (R), we start by analyzing each statement:

Assertion (A): The statement mentions that the net dipole moment of a polar linear isotropic dielectric is nonzero even without an external electric field. This assertion implies that there is a preferential alignment of the dipoles resulting in a non-zero net dipole moment in the absence of an external force, which is contrary to the basic understanding of polar dielectrics.

Reason (R): It states that without an external electric field, the permanent dipoles in a polar dielectric are oriented randomly. This aligns with our understanding that in such a case, the random order of dipoles cancels out each other's effects, leading to a net dipole moment of zero.

Hence, upon evaluation:

We conclude that option (A) is not correct but (R) is correct is the most appropriate. The assertion (A) incorrectly suggests a net dipole without an external field, while the reason (R) accurately describes the dipole behavior.

To solve this problem, we need to determine how the de Broglie wavelength of an electron changes over time when it is subjected to an electric field. Let's break this down step-by-step:

1. The initial de Broglie wavelength of the electron, , can be expressed as: , where is Planck's constant, is the mass of the electron, and is the initial velocity.
2. When the electron enters the electric field , it experiences a force: , where is the charge of the electron.
3. This force affects the electron's motion and results in an acceleration: .
4. The velocity of the electron after time can thus be given by: .
5. The speed of the electron after time is: .
6. The new de Broglie wavelength is then: .
7. We can express in terms of : .
8. Further simplification gives: .

This corresponds to the correct option: . This shows that the de Broglie wavelength decreases as the speed of the electron increases due to the electric field component in the direction, thereby validating the selected option.