Two point charges of 1 microcoulomb and -1 microcoulomb are separated by a distance of 100 Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be
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Given : Two point charges of 1 μC and -1 μC are separated by a distance of 100 A° . A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges.
To find : The electric field at P will be
solution : This question is based on electric field due to dipole.
we know electric field due to dipole at equatorial line is given by,
E = kP/(r² + l²)³/²
where l is the half of seperation between charges, P is dipole moment and r is the seperation between midpoint of charges and point of observation.
here P = q(2l)
= 1μC × 100 A°
= 10^-6 × 100 × 10^-10 Cm
= 10¯¹⁴ Cm
here r = 10cm = 0.1 m and l = 100/2 = 50A° = 5 × 10^-9 m
it is clear that, r >> l
so, E = kP/r³ , [ (r² + l²)³/² ≈ r³ ]
now E = (9 × 10^9 × 10¯¹⁴)/(0.1)³
= 9 × 10^-5/0.001
= 9 × 10¯² N/C
= 0.09 N/C
Therefore the Electric field at point P will be 0.09 N/C.
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