A rod and two balloons A very thin glass rod 4 meters long is rubbed all over wi

Question

A rod and two balloons
A very thin glass rod 4 meters long is rubbed all over with a silk cloth. It gains a uniformly distributed charge 1.1 106 C. Two small spherical rubber balloons of radius 1.2 cm are rubbed all over with wool. They each gain a uniformly distributed charge of -8 108 C. The balloons are near the midpoint of the glass rod, with their centers 3 cm from the rod. The balloons are 2 cm apart (4.4 cm between centers).

Use approximate formula for electric field of a charged spherical shell.Assume distance to observation location is small compared to length of rodNeglect polarization of balloonsNeglect polarization of rod

Explanation / Answer

part a:

as the rod's length is much greater as compared to dimensions of the balloons and their distance from the rod,

the rod can be approximated as an infinitely charged conductor.

in that case, linear charge density of the rod=total charge/length of the rod

=pho=1.1*10^(-6)/4=2.75*10^(-7) C/m

as the point marked X is inside the left ballon, their wont be any electric field due to

charge contained in left balloon (using gauss law, electric field density=charge enclosed by gaussian surface

but as due to left balloon, charge enclosed by a sphere through X is zero, there wont be any electric field

due to left balloon)

the right balloon can be taken as a point charge centered at its center

so distance of charge of -8*10^(-8) C contained on the right balloon from the point marked as X

=0.6 cm + 2 cm+1.2 cm=3.8 cm=0.038 m

electric field calculation:

field due to charged rod:

as the rod is positively charged, field will be directed away from the rod and towards X

as due to a charged rod , field is directed in perpendicular direction to the rod, direction of field here will be along -ve y axis

magnitude of field is given by linear chanrge density/(2*pi*epsilon*perpnedicular distance)

where epsilon=electrical permitivity of free space

perpendicular distance=3 cm=0.03 m

hence field due to charged rod=2.75*10^(-7)/(2*pi*8.85*10^(-12)*0.03)=1.648*10^5 N/C

in vector notation, field due to charged rod=-1.648*10^5 j N/C

where j is unit vector along +ve y axis

electric field due to the right sphere:

field due to a point charge of q is given by 9*10^9*q/d^2

where d is distance of the point where field to be calculated from the point charge

as here charge is negative, field will be directed towards the charge i.e. towards +ve x axis

magnitude of field=9*10^9*8*10^(-8)/0.038^2=4.986*10^5 N/C

in vector notation, field=4.986*10^5 i N/C

where i is unit vector along +ve x axis

hence total field in vector notation=4.986*10^5 i - 1.648*10^5 j N/C

part b:

force on a proton=charge*electric field

=1.6*10^(-19)*(4.986*10^5 i - 1.648*10^5 j)=7.9776*10^(-14) i - 2.6368*10^(-14) j N

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