Question

An electret is similar to a magnet, but rather than being permanently magnetized, it has a permanent electric dipole moment. Suppose a small electret with electric dipole moment 1.0 × 10−7 Cm is 25 cm from a small ball charged to + 80 nC, with the ball on the axis of the electric dipole.What is the magnitude of the electric force on the ball?

Answer 1

Answer:

 F = 9.216 × 10⁻³ N

Explanation:

given,

dipole moment = 1 × 10⁻⁷ Cm

distance apart from +80 nC charge = 25 cm

to calculate the magnitude of electric force

Electric field due to dipole

$E = \dfrac{2p}{4\pi \varepsilon_0 r^3 }$

$E = \dfrac{9\times 10^9\times 2\times 1 \times 10^{-7}}{25^3\times 10^{-6} }$

E = 1.152 × 10⁵ N/C

electric force on the ball

F = E q

  = 1.152 × 10⁵ × 80 × 10⁻⁹

 F = 9.216 × 10⁻³ N  

Hence, the electric force is equal to  F = 9.216 × 10⁻³ N

Answer 2

Answer:

Explanation:

Given that,

Chase Q = 80nC = 80×10^-9C

dipole moment p = 1×10^-7 Cm

Distance z = 25cm = 0.25

The Electric field along a dipole moment is given as

E = p / (2π•εo•z³)

We know that k = 1/4π•εo

Then, Electric field becomes

E = 2kp/r³

Where k = 9×10^9 Nm²/C²

Then, the force is given as

F = qE

Therefore,

F = 2k•q•p/r³

F = 2× 9×10^9 ×80×10^-9×1×10^-7/0.25³

F = 9.216 × 10^-3 N