(a)  Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

(b)  Two identical point charges, q each, are kept 2m apart in air. A third point        charge Q of unknown magnitude and sign is placed on the line joining the        charges such that the system remains in equilibrium. Find the position and         nature of Q.

 

 

 

 
 
 
 
 

Answers (1)

(a) Let us consider the two charges q and -q, be seperated by a distance 

       2a.

On the equatorial line of the dipole let any point P.at a distance r from the line joining the two charges.

Now,

    d=\sqrt{r^{2}+a^{2}}

The electric field at P due to the charge is : for q, the electric field :

    E_{q}=\frac{1}{4\pi \epsilon _{0}}\frac{q}{d^{2}}=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}+a^{2}}

For -q, the electric field :

    E_{-q}=\frac{1}{4\pi \epsilon _{0}}\frac{q}{d^{2}}=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}+a^{2}}

Adding horizontal components, such that vertical will cancel out.

Hence, 

    \vec{E}=(\vec{E_{q}}+\vec{E_{-q}})i\cos \theta

Let us suppose, \hat{P} be the unit vector in the direction of the dipole moment.

    \vec{E}=(\vec{E_{q}}+\vec{E_{-q}})\cos \theta\hat{P}

    E=2\; \frac{1}{4\pi \epsilon _{0}}\; \frac{q}{r^{2}+a^{2}}\cos \theta \hat{P}

    \cos \theta =\frac{a}{d}=\frac{a}{\sqrt{r^{2}+a^{2}}}

    \vec{E}=\frac{1}{4\pi \epsilon _{0}}\frac{2qa}{(r^{2}+a^{2})\frac{3}{2}}\hat{P}

    \vec{E}=\frac{1}{4\pi \epsilon _{0}}\frac{2\vec{P}}{(r^{2}+a^{2})\frac{3}{2}}

If, r> > a

Then, E=\frac{1}{4\pi \epsilon _{0}}\frac{2\vec{P}}{r^{3}}

(b) According to the Coulomb's law, All the charges must be in equilibrium and the force depends upon the distance and hence, the position of Q  must be in the mddle of the two charges.

Now, the nature of Q  must have opposite sign to that of q, to balance the forces.

 

 

 

 

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