(a) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.

(b) Derive an expression for the electric potential at any point along the axial line of an electric dipole.

 

 

 

 
 
 
 
 

Answers (1)

We know that the electric field strength is proportional to the distance between the surface, so, the surface will be evenly spaced.


Such that, the electric field is always perpendicular to the equipotential surface.

Hence, the equipotential surfaces are going to be infinite plane sheets in the z-direction parallel to the x-y plane.

(b) \Rightarrow Let us consider an electric dipole having -q and q charges separated by a certain distance (2a).

Now, we need to find the electric potential at any point along the axial line of an electric dipole.

 

Let the point 'P' along the axial line whose distance from the centre of the dipole is 'r'

Thus, the electric potential at 'p' due to charge A will be V_{1} and the electric potential at 'p' due to charge B will be V_{2}

Now, calculating the net potential at point 'p'

We have,

    V_{net}=V_{1}+V_{2}            \left [ \because v=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r} \right ]

    V_{net}=\frac{-1}{4\pi \epsilon _{0}}\frac{q}{(r+a)}+\frac{1}{4\pi \epsilon _{0}}\frac{q}{(r-a)}

    V_{net}=\frac{q}{4\pi \epsilon _{0}}\left [ \frac{1}{r-a}-\frac{1}{r+a} \right ]

    V_{net}=\frac{1}{4\pi \epsilon _{0}}q\left [ \frac{r+a-r+a}{r^{2}+a^{2}} \right ]        

V_{net}=\frac{1}{4\pi \epsilon _{0}}\frac{q(2a)}{r^{2}-a^{2}}

We know that, dipole moment  \vec{p}=2qa

    V_{net}=\frac{1}{4\pi \epsilon _{0}}\frac{p}{(r^{2}-a^{2})}

 For r> > a

    V_{net}=\frac{1}{4\pi \epsilon _{0}}\frac{p}{r^{2}}

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