Answer :
Answer:
Value of electric field along the axis and equitorial axis [tex]E=31.25\ N/c[/tex] and [tex]E = 15.625\ N/c[/tex] respectively.
Explanation:
Given :
Distance between charges , [tex]d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.[/tex]
Magnitude of charges , [tex]q=1\ nC = 10^{-9}\ C.[/tex]
Dipole moment , [tex]p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.[/tex]
Case A) (x,y) = (12.0 cm, 0 cm) :
Electric field of dipole in its axis ,
[tex]E=\dfrac{2kp}{r^3}[/tex]
Putting all values and [tex]r=12\times 10^{-2}\ m.[/tex]
We get , [tex]E=31.25\ N/c.[/tex]
Case B) (x,y) = (0 cm, 12.0 cm) :
Electric field of dipole on equitorial axis ,
[tex]E = \dfrac{kp}{r^3}[/tex]
Putting all values and [tex]r=12\times 10^{-2}\ m.[/tex]
We get , [tex]E = 15.625\ N/c.[/tex]
Hence , this is the required solution.