Electric energy stored in dipole

[mbmcgee]

there is a +q and -q on the x-axis. the +q is L/2 in the negative direction and the -q is L/2 in the positive directions. the distance between the two charges is L. there is a test charge P on the y-axis a distance r from both charges.

we had to find the electric energy stored in the dipole.

she gave us the answer -> W= (q^2)/(4*pi*epsilon_0*L) => W=(q_1*q_2)/(4*pi*r_12)

i do not understand how she got this. could someone please enlighten me a little bit?
thanks

 

[Hootenanny]

there is a +q and -q on the x-axis. the +q is L/2 in the negative direction and the -q is L/2 in the positive directions. the distance between the two charges is L. there is a test charge P on the y-axis a distance r from both charges.

we had to find the electric energy stored in the dipole.

she gave us the answer -> W= (q^2)/(4*pi*epsilon_0*L) => W=(q_1*q_2)/(4*pi*r_12)

i do not understand how she got this. could someone please enlighten me a little bit?
thanks

You can calculate the potential energy of the dipole the same way in which you calculate the potential energy of any array of charges.

 

[baba89]

I can explain how this equation for the electric energy stored in a dipole was derived. First, it is important to understand that electric energy is a form of potential energy, which is the energy an object has due to its position or configuration. In this case, the dipole is made up of two charges, +q and -q, separated by a distance L.

To calculate the electric energy stored in a dipole, we need to consider the work done to bring the charges together from infinity to their current positions. This work is equal to the potential energy of the dipole. The equation for potential energy is U = qV, where q is the charge and V is the potential. In this case, the potential is created by the two charges, so we need to consider the potential at the point P on the y-axis.

Using the equation for potential due to a point charge, V = (kq)/r, where k is the Coulomb's constant and r is the distance between the charge and the point P, we can calculate the potential at point P due to each of the charges in the dipole. Since the potential at point P is the sum of the potentials due to each charge, we can write it as V = V1 + V2, where V1 and V2 are the potentials due to +q and -q, respectively.

Next, we need to consider the work done to bring the test charge P to its current position, which is a distance r from both charges. This work is equal to the potential energy of the test charge, which we can calculate using the same equation U = qV. However, in this case, the potential is only due to one charge, so we can write it as U = qV1.

Now, the total potential energy of the dipole is the sum of the potential energies of the dipole and the test charge, so we can write it as Utotal = U + U1 = q(V1 + V2). Substituting in our equations for V1 and V2, we get Utotal = q((kq)/(r+L/2) + (kq)/(r-L/2)).

Finally, we can simplify this equation using the fact that q = +q - (-q) = 2q, and also substituting in the value of k, which is 1/(4*pi*epsilon_0). This gives us