Three identical dipoles are arranged as shown | vedclass

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(C) Point $P$ lies on the equatorial line of dipole $1$ and dipole $2$,and on the axial line of dipole $3$.For an equatorial point at distance $x$,the

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(C) Point $P$ lies on the equatorial line of dipole $1$ and dipole $2$,and on the axial line of dipole $3$.For an equatorial point at distance $x$,the electric field is $E_{eq} = \frac{kp}{x^3}$ directed opposite to the dipole moment.For an axial point at distance $x$,the electric field is $E_{ax} = \frac{2kp}{x^3}$ directed along the dipole moment.$1$. Due to dipole $1$: $E_1 = \frac{kp}{x^3}$ (directed towards the left).$2$. Due to dipole $2$: $E_2 = \frac{kp}{x^3}$ (directed towards the left).$3$. Due to dipole $3$: $E_3 = \frac{2kp}{x^3}$ (directed towards the right).The total electric field at $P$ is $E_{net} = E_3 - (E_1 + E_2) = \frac{2kp}{x^3} - \left( \frac{kp}{x^3} + \frac{kp}{x^3} \right) = \frac{2kp}{x^3} - \frac{2kp}{x^3} = 0$.

Three identical dipoles are arranged as shown below. What will be the net electric field at $P$ $\left( {k = \frac{1}{{4\pi {\varepsilon _0}}}} \right)$?

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