Suppose you have a dipole that's free to move in any way (including rotate - imagine it floating in space). And there's an objec
t with charge Q a distance r away. That distance r would be much larger than d, the distance between the charges of the dipole, so we draw the dipole small.
1 answer:
Complete Question
The complete question shown on the first uploaded image
Answer:
a)
The force on Q due to dipole is Attractive
b)
The charge Q exerts attractive force on the dipole
c)
Yes from the above parts, force depends on the sign of charge
d)
e)
The magnitude o force decrease by a factor of 8.0 times
Explanation:
The explanation is shown on the second uploaded image
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The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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