consider an electric dipole lying along x-axis with mid-point O as the origin of coordinate system. find the electric potential V at point P due to dipole. in addition to this, find electric potentials for special cases. when P lies on the axial line and when P lies on the equatorial line.

axial   V = 0

equatorial  V = k q 2a / (x² -a²),  V = k q 2x / (a² -x²)

Explanation:

A dipole is a system formed by two charges of equal magnitude, but different sign, separated by a distance 2a; let's look for the electrical potential in an axial line

       V = k (q / √(a² + y²) - q /√ (a² + y²))

        V = 0

the potential on the equator

we place the positive charge to the left and perform the calculation for a point outside the dipole

    V = k (q / (x-a) - q / (x + a))

    V = k q 2a / (x² -a²)

 we perform the calculation for a point between the dipo charges

     V = k (q / (a-x) - q / (a + x))

     V = k q 2x / (a² -x²)

the bal will not change its velocity.

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat