Answer:
1. 31.25 N/c
2. 15.625 N/c
Explanation:
Dipole moment of the dipole, p = q*d where d is the separation distance and q is charge
Substituting d with 3 mm which is 0.003 m and q with 1*10^{-9}
[tex]p= 1*10^{-9}*0.003= 3*10^-12 C.m[/tex]
A) at (x,y) = (12.0 cm, 0 cm)
so, at 12 cm on x axis
Electric filed due to dipole on its axis,
[tex]E = \frac {2*k*p}{r^{3}}[/tex] where r is x axis point in meters hence 0.12 m and taking k for axial line which is [tex]9*10^{9}[/tex]
[tex]E= {2*9*10^{9}*3*10^{-12}}{0.12^{3}}=31.25 N/c[/tex]
(b) at (x,y) = (0 cm, 12.0 cm)
Electric filed due to dipole on equatorial line,
[tex]E =\frac {k*p}{r^{3}}[/tex]
NB: This formula is different from the one in part (a) because here we consider the equatorial line
[tex]E= \frac {9*10^{9}*3*10^{-12}}{0.12^{3}}=15.625 N/c[/tex]
The electric field strength at the given point is 12 cm on the X-axis is 31.59 N/C.
A dipole is the arrangement of ions or particles with the same magnitude but different charges. The is the electric field strength in a dipole can be calculated by,
[tex]F = \dfrac {2kp}{r^3}[/tex]
Where,
[tex]k[/tex] - constant = [tex]9\times 10^9[/tex]
[tex]r[/tex] - distance = 12 cm
[tex]p[/tex] - dipole moment = [tex]3\times 10^{-12}[/tex]
Put the values in the formula,
[tex]F = \dfrac {2\times 9\times 10^9 \times 3\times 10^{-12}}{ 12^3}\\\\F = 31.59 \rm \ N/C[/tex]
Therefore, the electric field strength at the given point is 12 cm on the X-axis is 31.59 N/C.
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