Given two charges, (3C) and (-3C), separated by (1cm), and their distances from a point (P) being (5cm) and (7cm) respectively, we need to find the distance between their midpoint and the point (P).

First, let’s establish the positions of the charges and the point (P). Let’s assume the two charges are on a straight line, and (P) lies off this line. We are essentially working with a right-angled triangle, where the line segment connecting the charges acts as the base, and the lines connecting (P) to both charges act as the sides of the triangle. The midpoint of the charges will also lie on this base.

The charges are (1cm) apart. So, the midpoint of this (1cm) segment is (0.5cm) from each charge. This midpoint does not change the distances (P) has from the original positions of (5cm) and (7cm), as those lines would still emanate from the same points on the charges’ line, but we now consider the perpendicular from (P) to the midpoint.

Given the distances of (5cm) and (7cm), these seem to directly relate to the distances from (P) to each charge, not directly helpful in a simple calculation of the distance to the midpoint due to the geometry not being fully detailed. However, if we attempt to resolve this using the

When considering an electric dipole, consisting of two charges of equal magnitude but opposite sign separated by a distance (d), the potential due to this dipole at a point in space can be derived from the principle of superposition. The electric potential (V) at any point due to a single point charge is given by the equation:

[V = frac{kQ}{r}]

where:

– (V) is the electric potential,

– (k) is Coulomb’s constant ((8.987 times 10^9 Nm^2/C^2)),

– (Q) is the charge,

– (r) is the distance from the charge to the point in space where the potential is being calculated.

For a dipole, the total potential at any point is the sum of the potentials due to each charge. At a point (P) that is perpendicular to the midpoint of the dipole, you would be considering a point along the axis that is perpendicular to the line joining the two charges (equatorial line). At this point, the contributions to the potential from each charge in the dipole are equal in magnitude but opposite in direction, effectively cancelling out. Consequently, if the observation point (P) is equidistant from both charges, the potential (V) at point (P) due to the dipole is theoretically zero.

However, for a general position perpendicular to the dipole but not equidistant to both charges (say,

The dipole moment ((p)) is calculated using the formula: (p = q times d), where (q) is the magnitude of the charge and (d) is the separation distance between the charges.

Given:

– Charge, (q = 2C) (Coulombs)

– Separation distance, (d = 2 cm = 0.02 m) (since (1 cm = 0.01 m))

Therefore,

[p = q times d = 2C times 0.02m = 0.04 Ccdot m]

So, the dipole moment for the given configuration is (0.04 Ccdot m).

To compute the power consumed by the material, we can use the formula for power ((P)) in terms of the potential difference ((V)) and current ((I)), which is given by:

[P = V times I]

Here, (V = 20V), but (I) is not directly given. Instead, we have the current density ((J)) and area ((A)), which we can use to find (I), as the current density is defined as the current per unit area (J = frac{I}{A}). Rearranging this for (I) gives us:

[I = J times A]

Given:

– (J = 15) units (assuming the unit is Amperes per square meter, (A/m^2), for current density, which is a common unit),

– (A = 100) units (assuming square meters, (m^2), for area, which matches the unit for current density),

[I = 15 times 100 = 1500 A]

Now, using the formula for power:

[P = 20V times 1500A = 30,000 W]

Therefore, the power consumed by the material is 30,000 Watts or 30 kW.