Electrostatics

In a uniform electric field , the potential difference between two points in the field can be calculated using the equation , where is the distance in the direction of the field.

Given the electric field points in the positive X direction, let's evaluate potential differences:

• At Point O (Origin): Assume zero potential, .
• At Point A (r = +2 cm on X-axis): Since the electric field is along the X-axis, (convert to meters, ).
  The potential because distance in electric field direction decreases potential.
• At Point B (y = +1 cm on Y-axis): Since there's no movement along the field direction, there is no change in potential from origin. Therefore, .

Comparing potentials:

• and , so .
• , so .

Thus, the correct answer is .

Let's analyze the effect of introducing an insulating slab into a parallel plate capacitor.

1. Original Capacitance (C):

Let the original separation between the plates be d = 4 mm.

C = ε₀A / d

2. Capacitance with Insulating Slab:

When a slab of thickness t = 4 mm and dielectric constant K is introduced, the capacitance (C') becomes:

C' = ε₀A / [d - t + (t/K)]

3. Restoring Original Capacitance:

To restore the original capacitance, the distance between the plates is increased by 3.2 mm. Let the new separation be d'.

d' = d + 3.2 mm = 4 mm + 3.2 mm = 7.2 mm

The new capacitance (C'') is:

C'' = ε₀A / d'

We are given that C'' = C:

ε₀A / d' = ε₀A / d

d' = d

However, we are told to find the K value where the original capacitance is restored after inserting the dielectric, and increasing the distance to 7.2 mm. So:

ε₀A / d = ε₀A / [d - t + t/K], and we are told that to restore the original capacitance the distance needs to be increased to 7.2 mm. therefore:

d = d' - (d- (d - t + t/K))

d = d' - (d-d + t -t/K)

d = d' - (t-t/K)

4mm = 7.2mm - (4mm-4mm/K)

4 = 7.2 - 4 + 4/K

4 = 3.2 + 4/K

0.8 = 4/K

K = 4/0.8 = 5

Therefore, the dielectric constant of the material is 5.

The correct answer is:

Option 2: 5

An electric dipole in an electric field experiences forces and torques due to the interaction with the field. Let's analyze the scenarios:

Force (F) on the Dipole: The force on a dipole in a non-uniform electric field is generally not zero. This is because the strength of the electric field varies at different points in space, leading to a net force on the dipole.

Torque (τ) on the Dipole: Torque depends on the angle between the dipole moment (p) and the electric field (E). The torque τ is given by:

where θ is the angle between the dipole moment and the electric field. If the dipole moment is parallel to the electric field, θ is 0, making sin θ = 0. This results in:

Conclusion: For a dipole in a non-uniform electric field, with the dipole moment parallel to the field, the force F is non-zero (F ≠ 0), while the torque τ is zero (τ = 0).

Correct Answer:

To determine the forces experienced by a point charge placed at positions and between two large plane parallel sheets with equal but opposite surface charge densities and , we can use the principle of superposition and the characteristics of electric fields generated by infinite sheets. An infinite sheet with surface charge density produces a constant electric field perpendicular to the sheet. For two sheets with equal and opposite charges:

• At any point outside the region between the sheets, the fields from each sheet cancel each other, resulting in a net electric field .
• Within the region between the sheets, the fields add up, creating a constant electric field directed from the positive to the negative sheet.

Thus, we conclude that the forces experienced by the charge are , , , based on the conditions defined in such a setup with infinite sheets.

Capacitor System Solution:

Given:

• Two 40 μF capacitors in series
• One capacitor has dielectric (K=4)

Calculations:

1. Capacitor with dielectric:

2. Other capacitor: (unchanged)

3. Series combination formula:

4. Equivalent capacitance:

Correct Answer:

To solve this problem, we want to find the charge on the copper ball such that it remains suspended in the oil. This means that the net force on the ball is zero. The forces acting on the copper ball are:

1. Gravitational force acting downward:
2. Buoyant force acting upward:
3. Electrical force acting upward:

where:

• is the volume of the ball
• is the density of copper
• is the density of oil
• is the acceleration due to gravity
• is the charge on the ball
• is the electric field intensity

The volume of the sphere is given by:

where .

Thus,

.

The condition for suspension is that the net upward force equals the downward force:

.

Substituting the expressions for each force, we have:

.

Rearranging for ,

.

Substitute the given values:

Therefore,

.

This simplifies to:

.

.

Thus, the correct answer is: .

To solve this problem, we need to determine the ratios and where two capacitors of capacitance values 2 and 3 are in series, connected to a battery of V volts.

For capacitors in series, the total capacitance is given by:

Substituting the given values:

Solving for :

In a series, the charge across each capacitor is the same. Using , the voltage across each capacitor can be found as:

,

This implies .
The energy stored in a capacitor is given by:

Thus, .
Both ratios

and are , thus confirming the correct option:

The maximum induced EMF in a rotating coil is given by the formula:
ε_max = NABω
where:- N is the number of turns of the coil, A is the area of the coil in square meters, B is the magnetic field in Tesla, ω is the angular velocity in radians per second.

Given:

• Electric field N/C (where )
• Cube side length cm m
• Cube is placed at cm m from the origin
• Electric field is directed along the positive -axis

Step 1: Determine the electric field at each face

The cube has two faces perpendicular to the -axis:

• Left face at m (since the cube is placed 1 cm from the origin)
• Right face at m (since the cube has side length 1 cm)

Electric field values:

• At left face: N/C
• At right face: N/C

Step 2: Calculate the flux through each face

Flux through a face is given by , where is the angle between and the normal to the face.

For the left face:

• Area m²
• (since points right and the normal to the left face points left)
• Wb

For the right face:

• Area m²
• (since and the normal both point right)
• Wb

For the other four faces (parallel to the -axis), the flux is zero because is perpendicular to their normals.

Step 3: Calculate the net flux

Net flux

Wb

The equation for charge is:

where:

• is in coulombs (C)
• is time in seconds (s)

The initial value of current.

Current ( ) is defined as the rate of change of charge with respect to time:

Differentiating the given charge equation:

The initial current is the current at time :

The initial value of current is .

Let the charges and be fixed on a line with a separation of 60 cm. We are to find a location to place a third charge (say ) such that the net electrostatic force acting on it is zero.
We consider three possible regions to place the third charge:
(i) to the left of , (ii) between and , and (iii) to the right of .

Case (ii): Between and
Assume the test charge is placed at a point between the two charges. Since both charges will exert attractive or repulsive forces in the same direction (depending on the sign of ), the forces will not cancel.
So, equilibrium cannot occur between the charges.

Case (i): To the left of
Let the third charge be placed at a distance cm to the left of . Then its distance from will be cm.
Using Coulomb's law, equating magnitudes of forces for net force to be zero:

So, the third charge should be placed 60 cm to the left of . This matches Option (C) since the point is 60 cm from . Let’s confirm directions:
- Force due to is repulsive.
- Force due to is attractive.
Both act in opposite directions if the test charge is to the left of , and they can cancel only in this region.

To solve this problem, we'll use Coulomb's Law, which states the force between two point charges and separated by a distance is given by:



where is Coulomb's constant. Initially:

Now, the distance is doubled, and both charges are halved, so the new distance is and the new charges are and . The new force can be expressed as:





Thus, the new force compared to the original force is:




Therefore, the new force is .

To find the power dissipated in a resistor, we use one of the standard formulas from electric power concepts. The three most common formulas for power are:
1.
2.
3.
Here, we are given:
Current
Resistance
We are not given the voltage, so the most convenient formula to use is:
Step 1: Square the current
Step 2: Multiply by resistance
Interpretation: The resistor converts 40 joules of electrical energy into heat every second. That’s what we mean by “dissipated power” — energy that’s lost (usually as heat) due to resistance.
Alternative :
We can also find the voltage across the resistor using Ohm’s law:

Now, apply :

This confirms the same result.