Masterinc Physics Chapter 23- Electric d Google chrome Secure l https:// on.mast

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Masterinc Physics Chapter 23- Electric d Google chrome Secure l https:// on.masteringphysi 74942428 Intro E & M Problem 23.50 Problem 23.50 Part A A dipole is centered al the origirl of a coor dinale systern, arr d a smal he dipole charge s 12 nC and the dipole separation is 20 mm, how tar away trom the dipole s the sphere? charged sphere is some distance avay along the perpendicula Express your answer with the appropriate units. bisector of the dipole. The particle carries a uniformly distributed charge of -4.0 IC arid experiences a 290 LN electric force in the pasir y direction alue Units Submit My Answers Up Incorrect; Try Again: 6 attempts remaining Part B How is the dip Orienleu? The dipole moment points in the l y direction. The dipole moment points in the +r dirertion pull dipole a diredion. The dipole moment points in the y direction. Submit Up Correct 4 at Continue 28 AM 1/29/2017 R6

Explanation / Answer

we know that , electric field due to dipole at any point along the perpendicular bisector

is directed in a direction opposite to that of the dipole moment.

here a negative charge experiences a force in +ve y direction

as force=charge*electric field

direction of force on a negative charge is opposite to direction of field

so direction of electric field is in -ve y direction

then direction of dipole moment is in +ve y direction

magnitude of electric field of a dipole at a point along the perpendicular bisector is given by

2*k*q*a/(x^2+a^2)^(1.5)

where k=coloumb's constant=9*10^9

q=dipole charge magnitude

a=half of distance between the dipoles

x=distance of the charge on which force is exerted

as per the question,

electric field magnitude=force/charge

=290*10^(-9)/(4*10^(-9))=72.5 N/C

a=20 mm/2=10 mm=0.01 m

q=12 nC=12*10^(-9)

x to be calculated.

then 2*k*q*a/(x^2+a^2)^1.5=72.5

==>2*9*10^9*12*10^(-9)*0.01/(x^2+0.01^2)^1.5=72.5

==>(x^2+0.0001)^1.5=0.029793

==>x^2+0.0001=0.0961

==>x^2=0.096

==>x=0.31 m

hence the charge is at a distance of 0.31 m from the dipole

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