Question

Two points A and B are located at distance of 100 cm from on electric dipole. A is sn axial point while B is in perpendicular bisector. The electric field intensity at A is $\overrightarrow{E}$. Then the intensity at B will be

• A. $\overrightarrow{E}$
• B. $2\overrightarrow{E}$
• C. $-\overrightarrow{E}$
• D. $-\overrightarrow{\frac{E}{2}}$

Answer 1

The correct option is D $-\overrightarrow{\frac{E}{2}}$

Two points $A$ and $B$ are located at distance $=100cm=100\times {10}^{-2}m$

The electric field intensity at $A=\overline{E}$

Then, the intensity at $B$ will be

$\frac{E_1}{F}\frac{E_A}{E_B}=\frac{100}{50}$

$\Rightarrow E_B=\frac{1}{2}{E}_A$