Question

Find electric field intensity due to dipole of dipole moment
${4.4 \times 10}^{ - 19} c - m$
at a point 30nm away along bisector axis?​

Answer 1

Answer:

electric field will be=0.132

Answer 2

we have to find electric field intensity due to dipole of dipole moment 4.4 × 10^-19 C.m at a point 30nm away along bisector axis.

solution : electric field intensity due to dipole to dipole is given by, E = Kp/r³

where K = 9 × 10^9 Nm²/C²

dipole moment, p = 4.4 × 10^-19 C.m

r = 30nm = 30 × 10^-9 m

now E = (9 × 10^9 × 4.4 × 10^-19)/(30 × 10^-9)³

= (39.6 × 10^-10)/(27000 × 10^-27) N/C

= 1.47 × 10¹⁴ N/C

Therefore the electric field intensity due to dipole to dipole is 1.47 × 10¹⁴ N/C.

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