Question

two charges of +25nC and -25nC are placed 6m apart. find the intensity of electrical field at a point 4m from the centre of electrical dipole?
a) on axial line
b) On equatorel line​

Answer 1

${\huge{\underline{\orange{ANS}\green{WER}}}}$

$\bigstar$ $\sf{\green{\underline{\underline{\orange{CONCEPTS:-}}}}}$

(☞ Electric field due to dipole;

1. at axial point

• E = ${\dfrac{2\:k\:p}{r^3}}$

1. at equitorial point

• E = ${\dfrac{k\:p}{r^3}}$

→ where

• p = electric dipole moment = “charge(q) × distance between two charge(d)”
• k = ${\dfrac{1}{4\:\pi\:\epsilon_o}}$ = 9×10$^9$ N. m²/C²
• r = given distance on axial or equitorial point from center of electrical dipole .

(☞ Electric dipole moment (p);

• p = (q × d)

$\bigstar$ $\sf{\green{\underline{\underline{\orange{Given:-}}}}}$

• Two charges of “+25nC” and “-25nC” are placed 6m apart .
• So, ${q_1\:=\:+25nC}$
• ${q_2\:=\:-25nC}$
• distance between two charge (d) = 6m
• r = 4m

$\bigstar$ $\sf{\green{\underline{\underline{\orange{To\:Find:-}}}}}$

• The intensity of electric field at a point 4m from the center of electrical dipole,

1. On axial line.
2. On equitorial line.

$\bigstar$ $\sf{\green{\underline{\underline{\orange{SOLUTION:-}}}}}$

(☞ First find the electrical dipole moment (p);

${p\:=\:q\:\times\:d}$

$\tt{\implies\:p\:=\:(25\:\times\:10^{-9})\:\times\:6}$

$\tt{\implies\:p\:=\:(15\:\times\:10^{-8})\:C-m}$

(☞ Now calculate the intensity of electric field at point 4m from center of electrical dipole

1) AT AXIAL POINT:-

${E\:=\:{\dfrac{2\:k\:p}{r^3}}}$

$\tt{\implies\:E\:=\:{\dfrac{2\:\times\:9\times10^{9}\:\times\:15\times10^{-8}}{4^3}}}$

$\tt{\implies\:E\:=\:42.18\:N/C}$

2) AT EQUITORIAL POINT:-

${E\:=\:{\dfrac{k\:\times\:p}{r^3}}}$

$\tt{\implies\:E\:=\:{\dfrac{9\times10^{9}\:\times\:15\times10^{-8}}{4^3}}}$

$\tt{\implies\:E\:=\:21.09\:N/C}$

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:42.18\:N/C\:\:and\:\:21.09\:N/C\:}}}}$