Module 2 : Electrostatics
Lecture 9 : Electrostatic Potential
  Solution :
Consider the dipole shown above, where the charges are separated by a distance $\delta z$, so that the dipole moment is $p= q\delta z$. Let the field at the charge $-q$be $E$and that at be $E + \delta E$. The force on the charge $-q$is $-qE$and on $+q$is $q(E+\delta E)$. There is a net force $q\delta E$to the right (z-direction). We can write the force as
 
\begin{displaymath}{\rm Force } = q\delta E = q\delta z \frac{\delta E}{\delta z}= p\frac{dE}{dz}\end{displaymath}
  where the last equality is valid for an ideal dipole for which $\delta z\rightarrow 0$.
In our case $p= qd$and
 
\begin{displaymath}\frac{dE}{dz} = E_0\alpha\end{displaymath}
  so that the net force is $ qdE_0\alpha$.

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