# Variation of electric field with volume of dielectric in space

Suppose there are 2 spheres which have charges $+q$ and $-q$ and separated by a distance d. Let the field be $\vec{E_0}$.

Now I bring a dielectric rod which is eqal in length to the separation between the spheres and place it along the line joining their centres. Charges with opposite polarity are induced on the dielectric, ie, positive charge is induced on the end closer to the negative sphere and vice versa. The additional field produced be $\vec{E'}$. Since $\vec{E'}$ and $\vec{E_0}$ are along the same direction, the net field is greater in magnitude than $E_0$.

Now consider the spheres in the same configuration but which are now placed in a dielectric medium with the same dielectric constant as the rod. The net field in this case is lesser than $E_0$.

My question is, when does the electric field rise to greatest value? Initially, in the case of the rod, the field increased. But when the space was filled with the dielectric, the field decreased. What is the variation of $E$ with the space filled by the dielectric?

Update: I don't need a quantitative answer since the involved math may become too difficult. I just want to know under what conditions would the field start reducing in magnitude.

My guess: I think that the field would start reducing when both charges are enclosed in a volume of a dielectric like a cylinder surrounding the charges.If only one charge were to be enclosed, it seems to me that the charges would experience a greater force.

• I think your statement "are along the same direction, the net field is greater in magnitude" is incorrect. The fields will be in opposite directions. – jim Mar 28 '16 at 11:02
• a similar example which i have used can be found here (Fig 6.6.2). it is clearly stated that the applied field is lesser in magnitude than the new value of the field. – Akshit Mar 28 '16 at 11:41
• can you copy and paste the relevant statement? – jim Mar 28 '16 at 11:46
• "As the array of spheres is inserted between the electrodes, surface charges are induced... Nevertheless, the average field at the electrode is larger than the applied field $E_a$." – Akshit Mar 28 '16 at 11:47

The electric field generated by the polarized dielectric directly correlates to the polarization density of the dielectric. You can find the polarization density of the dielectric with: $$P=\epsilon_0X_eE$$ where $\epsilon_0$ is the electrical permittivity of free space, $X_e$ is the electrical suscepribility of the dielectric which depends on the dielectric material and $E$ is the external electric field which is the electric field of the spherical capacitor in this case. Then, you can find the amount of charge accumulated on either side of the rod by taking the closed surface integral of the object made out of insulating material; $$-Q=\iint_SPdA$$ Then, you can apply the integral form of the Gauss' Law and find the electric field due to polarization. You can alter the parameters and obtain the highest value for the magnitude of the electric field due to polarization.