The distance between the charges -1 - C and -1 - C of an electric dipole is 4 - 10-14m- Calculate the potential at an axial point 2 - 10-6m from the centre of the dipole-

[ |q|=1 μC=10^ 6C; 2 l=4 × 10^ 14m; 14 πε_0=9 × 10^9N m^2C^2 ] here r^2 l^2 Electric potential on the axis of dipole [ ...

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Byju's Answer

Standard XII

Physics

Potential at a General Point Due to a Dipole

The distance ...

Question

The distance between the charges $- 1 μ C$ and $+ 1 μ C$ of an electric dipole is $4 \times 10^{- 14} m$ . Calculate the potential at an axial point $2 \times 10^{- 6} m$ from the centre of the dipole.

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Solution

$\begin{matrix} | q | = 1 μ C = 10^{- 6} C 2 l = 4 \times 10^{- 14} m \frac{1}{4 π ε_{0}} = 9 \times 10^{9} \frac{N - m^{2}}{C^{2}} \end{matrix}$
here $r^{2} ⋙ l^{2}$
Electric potential on the axis of dipole
$\begin{matrix} V = \frac{1}{4 π ε_{0}} \frac{p}{r^{2}} V = \frac{1}{4 π ε_{0}} \frac{(q) \times 2 l}{r^{2}} V = \frac{9 \times 10^{9} \times 10^{- 6} \times 4 \times 10^{- 14}}{2 \times 10^{- 6} \times 2 \times 10^{- 6}} V = \frac{9 \times 10^{9} \times 10^{- 20}}{10^{- 12}} V = 9 \times 10^{21 - 20} V = 90 volt \end{matrix}$

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The distance between the charges −1μC and +1μC of an electric dipole is 4×10−14m. Calculate the potential at an axial point 2×10−6m from the centre of the dipole.

|q|=1μC=10−6C2l=4×10−14m14πε0=9×109N−m2C2here r2⋙l2Electric potential on the axis of dipoleV=14πε0pr2V=14πε0(q)×2lr2V=9×109×10−6×4×10−142×10−6×2×10−6V=9×109×10−2010−12V=9×1021−20V=90 volt

The distance between the charges $- 1 μ C$ and $+ 1 μ C$ of an electric dipole is $4 \times 10^{- 14} m$ . Calculate the potential at an axial point $2 \times 10^{- 6} m$ from the centre of the dipole.