Electric Field

In a uniform electric field, the electric potential decreases in the direction of the field. Since point B is at a lower potential than point A (due to the perpendicular direction of the line AB to the field lines), the potential at point B will be less than at point A. As point C lies along the field lines, the potential at C will be the lowest among the three. Therefore, the potential decreases as we move from A to B to C.

The correct option is (B) : V_A=V_B > V_C

For an infinitely long wire with uniform linear charge density , the electric field at a distance from the wire is given by the formula: Where: - is Coulomb's constant, - is the charge density, - is the distance from the wire. Substituting the values into the formula: Thus, the magnitude of the electric field at a distance 20 cm from the wire is .

Problem Analysis:

1. The problem provides a diagram of electric field lines.
2. It states that the separation between the field lines on the left (region B) is twice the separation between the field lines on the right (region A). Let the separation at A be and at B be . So, .
3. The magnitude of the electric field at point A is given as .
4. We need to find the force on a charge placed at point B.

Key Concept:

The density of electric field lines (number of lines per unit area perpendicular to the lines) is proportional to the magnitude of the electric field strength. Therefore, where the lines are closer together, the field is stronger, and where they are farther apart, the field is weaker. Specifically, the electric field strength is inversely proportional to the separation between the lines (assuming a uniform distribution in the perpendicular dimension shown):

This implies that the ratio of electric fields at two points is inversely proportional to the ratio of the separations at those points:

Calculations:

1. Use the relationship between field strength and separation:

Substitute :

1. Calculate the electric field strength at B ( ):

Given :

1. Calculate the force on the charge at point B ( ) using the formula :

Substitute and (Note: ):

The final answer is (A): .

Step 1: Recall the formula for the electric field due to a uniformly charged infinite wire

For an infinitely long, straight uniformly charged wire, the electric field at a point at a distance from the wire is given by Gauss's law.

Gauss's law states that:

where is the charge enclosed by the Gaussian surface, and is the permittivity of free space.

For an infinitely long wire, the electric field at a distance from the wire is radial and uniform in magnitude at all points on a circle around the wire.

Step 2: Use symmetry and apply Gauss's law

Consider a Gaussian surface in the form of a cylinder with its axis coincident with the wire. The radius of this cylindrical surface is , and the length of the cylinder is .

The electric flux through the curved surface of the cylinder is:

Since the electric field is uniform and radial, the flux through the two ends of the cylinder is zero. The charge enclosed by the Gaussian surface is the charge on the length of the wire, which is , where is the linear charge density.

By Gauss's law:

Step 3: Solve for the electric field

Solving for :

Thus, the electric field due to an infinitely long uniformly charged wire is inversely proportional to the distance from the wire.

Final Answer:
The electric field due to an infinite, straight uniformly charged wire varies with distance as , which matches option (B).

Given problem:

• Electric field at a distance from an infinitely long uniformly charged straight conductor with linear charge density is .
• Another uniformly charged conductor with the same linear charge density is bent into a semicircle of radius . The electric field at its center is .

We need to find the relationship between and .

Step 1: Electric Field due to an Infinitely Long Uniformly Charged Straight Conductor

The electric field at a distance from an infinitely long uniformly charged straight conductor with linear charge density is given by:

Step 2: Electric Field due to a Semicircular Conductor

Consider a semicircular conductor of radius with linear charge density . We need to find the electric field at the center of this semicircle.

To do this, we can use the principle of symmetry and integration. The semicircle can be divided into small segments, each contributing to the electric field at the center. Due to symmetry, the horizontal components of the electric field from each segment will cancel out, leaving only the vertical components.

The electric field due to a small segment of the semicircle at an angle from the vertical is given by:

where and .

Thus,

The vertical component of is:

Integrating from to :

Step 3: Comparing and

From the calculations, we see that:

Thus, .

Therefore, the correct answer is: