To find the distance between the midpoint of the two charges and the point P, we first understand that the charges are separated by 1 cm, implying they lie on a straight line with each other. The charges are 3C and -3C, but this information is not directly relevant for calculating the distance to the midpoint; it is more relevant in electrostatic force or field calculations.

Given distances from P:

– The distance from P to the 3C charge is 5 cm.

– The distance from P to the -3C charge is 7 cm.

Since the charges are 1 cm apart, if we represent this linearly, we have two scenarios depending on the order of the charges regarding point P:

### 1. P closer to 3C charge:

– P —– 5cm —– 3C —– 1cm —– -3C —– 7cm (not possible based on the given distances; this arrangement would mean P is closer to -3C)

### 2. P closer to -3C charge, which fits the given distances:

– 3C —– 1cm —– -3C —– 5cm —– P —– 7cm (also not logical based on initial direct reading but helps picturing the scenario opposite to the initial assumptions)

### Correct logical arrangement:

– Considering the midpoint between charges is directly in the middle of the 1 cm gap, making the distances from P to each charge conceptually misunderstood in the depiction.

To find the angle at which the potential due to a dipole is measured, given the distances from the charges, consider a dipole consisting of charges (+q) and (-q), separated by a distance (2a). The position where the potential is being measured forms a triangle with the line joining the charges. Given that the distances from the positive and negative charges are 12 cm and 11 cm, respectively, and the separation between the charges ((2a)) is 2 cm, we can find the angle using the law of cosines.

Let us denote:

– (r_+) as the distance from the positive charge ((12 cm)),

– (r_-) as the distance from the negative charge ((11 cm)),

– (d = 2 cm) as the distance between the charges,

– (theta) as the angle the position vector (from the negative to the positive charge) makes with the perpendicular bisector of the dipole.

Using the law of cosines on the triangle formed by the position of measurement and the two charges, we have:

[r_+^2 = r_-^2 + d^2 – 2 cdot r_- cdot d cdot cos(theta)]

Substituting values:

[12^2 = 11^2 + 2^2 – 2 cdot 11 cdot 2 cdot cos(theta)]

S

To compute the power consumed by the material, we can use the formula for power ((P)) in electrical systems, which is given by (P = IV), where (I) is the current in amperes (A) and (V) is the potential difference across the material in volts (V).

However, to find the current ((I)) flowing through the material, we first need to determine the total current from the given current density ((J)). Current density is defined as the current flow per unit area, given by (J = I/A), where (J) is the current density, (I) is the total current, and (A) is the area through which the current flows.

Given:

– Current density ((J)) = 15 units (assuming units are (A/m^2) for the purpose of this calculation, as actual units were not specified),

– Area ((A)) = 100 units ((m^2), assuming meters squared for coherence with current density units),

– Potential difference ((V)) = 20V.

We find the total current flowing ((I)) first:

[J = frac{I}{A} implies I = J times A]

[I = 15 , A/m^2 times 100 , m^2 = 1500 , A]

Then, we substitute (I) into

To find the current in the conductor, we need to understand the relationship between the given quantities. The electric field (E) in a conductor where an electric potential (V) is applied across a distance (d) can be expressed as (E = V / d). However, the electric field is given as 2 units (assuming SI units, this would be volts per meter), and the distance is 100 cm (which is 1 meter). From this, we can directly find the potential difference (V) across the conductor since (V = E times d).

Given:

– Electric field (E) = 2 V/m

– Distance (d) = 100 cm = 1 m

First, let’s find the potential difference (V):

[ V = E times d = 2 , text{V/m} times 1 , text{m} = 2 , text{V} ]

Now, Ohm’s law states that (V = IR), where (V) is the voltage across the conductor, (I) is the current through the conductor, and (R) is the resistance of the conductor. Given the resistance (R) is 2 ohms, we rearrange Ohm’s law to solve for the current (I):

[ I = frac{V}{R} ]

Substituting the given values:

[ I = frac{2 ,

To find the inductance of a coil, we can use the formula for the inductance of a solenoid, which is given by:

[L = mu N^2 A / l]

where:

– (L) is the inductance,

– (mu) is the permeability of the core material,

– (N) is the number of turns,

– (A) is the cross-sectional area,

– (l) is the length of the coil.

Given in the question:

– (mu = 3.5) (Assuming the unit here is Henrys per meter (H/m) since it’s about permeability and no specific unit is given),

– (N = 100) turns,

– The length ((l)) of the coil is (2m),

– The area ((A)) is thrice the length. However, to calculate area from length doesn’t directly compute without additional context, as length and area are not directly convertible. Assuming it means the side length of a square cross-section is thrice some basic unit, or perhaps a misunderstanding in the question’s phrasing. If it meant the area is proportional to the length in some manner not clearly defined, clarification is needed. Assuming a more typical approach where we define an area based off given or assumed dimensions: if by “thrice the length”, it means for example, the dimension contributing to the area is 3 times some value