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## Discuss and explain the relation between electric field intensity and potential

The relationship between electric field $E$ and scalar potential $ϕ$ is given as:

$E=▽ϕ$, where $▽=$gradient operator.

$E=▽ϕ$, where $▽=$gradient operator.

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The relationship between electric field $E$ and scalar potential $ϕ$ is given as:

$E=▽ϕ$, where $▽=$gradient operator.

$E=▽ϕ$, where $▽=$gradient operator.

2

Potential due to a dipole is given by:

$V≈4πε_{o}r_{2}pcosθ $

where

$p:$ electric dipole moment

$θ:$ angle made by the line joining center of dipole and the point with the dipole moment vector

$r:$ distance between the center of dipole and the point

Note:

Approximation is made that the length of dipole is negligible as compared to the distance of the point from the dipole.

$V≈4πε_{o}r_{2}pcosθ $

where

$p:$ electric dipole moment

$θ:$ angle made by the line joining center of dipole and the point with the dipole moment vector

$r:$ distance between the center of dipole and the point

Note:

Approximation is made that the length of dipole is negligible as compared to the distance of the point from the dipole.

3

Surfaces having same potential are termed as equipotential surfaces

The properties of equipotential surfaces can be summarized as follows:

The properties of equipotential surfaces can be summarized as follows:

- The electric field lines are normal to the equipotentials and are directed from higher to lower potentials.
- By symmetry, the equipotential surfaces produced by a point charge form a family of concentric spheres, and for a constant electric field, a family of planes normal to the field lines.
- The tangential component of the electric field along the equipotential surface is zero, otherwise non-vanishing work would be done to move a charge from one point on the surface to the other.
- Work done in moving a particle along an equipotential surface is zero.

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We know that $C=ΔVQ $

But $Q=σA$ and $ΔV=Ed$

putting these values in capacitance formula we get

$C=dϵ_{o}A $

Where, A = Area of plate, d = Distance between capacitor plates

But $Q=σA$ and $ΔV=Ed$

putting these values in capacitance formula we get

$C=dϵ_{o}A $

Where, A = Area of plate, d = Distance between capacitor plates

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Example:

In the figure shown, find the effective capacitance across P and Q. (Area of each plate is $α$)

Solution:

The above arrangement acts as the $K_{1}$ capacitor in parallel with $K_{2}$ and $K_{3}$ which are in series.

$C_{1}=2dk_{1}ε_{0}α ,C_{2}=2dk_{2}ε_{0}α2 ,C_{3}=2dk_{3}ε_{0}α2 $

$C_{2}$ and $C_{3}$ are in series.

$C_{eff}1 =C_{2}1 +C_{3}1 $

$C_{eff}1 =k_{2}ε_{0}αd +k_{3}ε_{0}αd =C_{eff}=dε_{0}α (k_{2}+k_{3}k_{2}k_{3} )$

$C_{1}$ is in parallel with $C_{eff}$

$∴$ Total capacitance $C=2dk_{1}ε_{0}α +dε_{o}α (k_{2}+k_{3}k_{2}k_{3} )$

$C=dε_{0}α (2k_{1} +k_{2}+k_{3}k_{2}k_{3} )$

In the figure shown, find the effective capacitance across P and Q. (Area of each plate is $α$)

Solution:

The above arrangement acts as the $K_{1}$ capacitor in parallel with $K_{2}$ and $K_{3}$ which are in series.

$C_{1}=2dk_{1}ε_{0}α ,C_{2}=2dk_{2}ε_{0}α2 ,C_{3}=2dk_{3}ε_{0}α2 $

$C_{2}$ and $C_{3}$ are in series.

$C_{eff}1 =C_{2}1 +C_{3}1 $

$C_{eff}1 =k_{2}ε_{0}αd +k_{3}ε_{0}αd =C_{eff}=dε_{0}α (k_{2}+k_{3}k_{2}k_{3} )$

$C_{1}$ is in parallel with $C_{eff}$

$∴$ Total capacitance $C=2dk_{1}ε_{0}α +dε_{o}α (k_{2}+k_{3}k_{2}k_{3} )$

$C=dε_{0}α (2k_{1} +k_{2}+k_{3}k_{2}k_{3} )$