Question

Kindly solve this if u know.

Answer 1

Let $\vec{\sf{p_1}},\ \vec{\sf{p_2}}$ and $\vec{\sf{p_3}}$ be the dipole moment in the upper dipole, right dipole and lower dipole respectively.

Since each dipole contain same magnitude of charges, the magnitude of each dipole moment should be the same.

• $\sf{p_1=p_2=p_3=p=2Qa}$

where $\sf{2a}$ is the length of the dipole.

We see negative charge is at left to the positive charge in each dipole. Hence the direction of dipole moment is same in each dipole and is rightwards, from negative to positive.

• $\sf{\hat{p_1}=\hat{p_2}=\hat{p_3}=\hat i}$

P is along equatorial line of the upper dipole. So the electric field due to this dipole at P will be,

$\longrightarrow\vec{\sf{E_1}}=\sf{-\dfrac{kp}{x^3}\ \hat i}$

P is along axial line of the right dipole. So the electric field due to this dipole at P will be,

$\longrightarrow\vec{\sf{E_2}}=\sf{\dfrac{2kp}{x^3}\ \hat i}$

P is along equatorial line of the lower dipole. So the electric field due to this dipole at P will be,

$\longrightarrow\vec{\sf{E_3}}=\sf{-\dfrac{kp}{x^3}\ \hat i}$

So the net electric field will be,

$\longrightarrow\vec{\sf E}=\vec{\sf{E_1}}+\vec{\sf{E_2}}+\vec{\sf{E_3}}$

$\longrightarrow\vec{\sf{E}}=\sf{-\dfrac{kp}{x^3}\ \hat i+\dfrac{2kp}{x^3}\ \hat i-\dfrac{kp}{x^3}\ \hat i}$

$\longrightarrow\underline{\underline{\vec{\sf{E}}=\sf{0}}}$

Hence (3) is the answer.