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

Step-by-step 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

Answer 2

F = 9.216 × 10⁻³ N

Step-by-step 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 = (2p)/(4\pi \varepsilon_0 r^3 )`

`E = (9* 10^9* 2* 1 * 10^(-7))/(25^3* 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