Physics, asked by ojasvisood3240, 1 year ago

Compare the field intensity due to an electric dipole when field intensity due to a simple charge for a given point

Answers

Answered by karan4185
0

The full amount to be circle

Answered by ғɪɴɴвαłσℜ
27

\Huge\bf\purple{\mid{\overline{\underline{Answer}}}\mid}

Electric Field Intensity :

The electric field intensity of any point is defined as the force experienced by unit positive charge placed at that point.

 \large\tt\red{E=\frac{F}{q_0}}

The SI unit of electric field intensity

 \large \tt\orange{NC^{-1}}

Electric field at any equatorial point of a dipole

 \large \tt{}is  \:  \:  \: \green {\vec{E}=\frac{1}{4\pi\:\varepsilon_0}\frac{\vec{p}}{(r^2+a^2)^\frac{3}{2}}}

At the mid-point of the dipole, r=0 so ,

\large \tt \purple{\vec{E}=\frac{1}{4\pi\:\varepsilon_0}\frac{\vec{p}}{a^3}}

\huge{\mathfrak{\orange{hope\; it\; helps}}}

Similar questions