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.

Answer:In this question, we need to find the magnitude of the electric force on the ball; the answer is [tex]9.21[/tex][tex]*[/tex][tex]10^{-3}[/tex]N.Step-by-step explanation:To resolve this question, we're going to need E=[tex]\frac{k(2p)}{r3 }[/tex] that is the magnitude of an electric field in the axis of a dipole at some distance "r" .where "p" is the magnitude of the electric dipole moment, so in this case p= 1.0[tex]*[/tex][tex]10^{-7}[/tex] Cm or what is the same [tex]10^{-7}[/tex]cm; and k is a constant k value is = [tex]k=8.99[/tex][tex]*[/tex][tex]10^{9}[/tex] [tex]\frac{Nm2}{C2}[/tex]now to get the force on the ball we need to use F=qE where [tex]E=[/tex][tex]\frac{k(2p)}{r3 }[/tex] so we have[tex]F= q*[/tex][tex]\frac{k(2p)}{r3 }[/tex]for this, "q" is the charge on the ball, so we got 80nC or what si the same 80×[tex]10^{-9}[/tex]C, and r is the distance between the ball and the dipole, r= 25cm= 0.25m (1m=100cm, so we need to divide 25cm into 100 to get it in meters) Replacing we got:F=(80×[tex]10^{-9}[/tex])(8.99×[tex]10^{9}[/tex])(2([tex]10^{-7}[/tex])) / [tex](0.25)^{3}[/tex]and calculating this the answer is [tex]9.21[/tex][tex]*[/tex][tex]10^{-3}N[/tex]