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Unformatted text preview: h direction the force will be
acting.
For D:
If the spheres are allowed to touch, some of the charge will cancel, leaving +5 μC to be
equally distributed between the two spheres. (i.e. +2.5 μC per sphere.)
U= q1 q 2
4πε 0 r ⎛
Nm 2 ⎞ (2.5μC )(2.5μC )
⎟
U = ⎜ 9 × 10 9
⎜
0.6cm
C2 ⎟
⎝
⎠
U = 9.375 J Solutions will be posted at www.prep101.com/solutions 19 ©Prep101 Phys1026 April Exam Solutions Question 9:
Solution:
For A:
Charge on a conducting sphere distributes itself evenly on the surface. For B:
Since the sphere has a radius of R, this is the electric field at a distance of 2R from the
center of the sphere. The conducting sphere can be treated as a point charge at the middle
of the sphere for the purpose of determining the electric field outside of the sphere.
E=
E= kq
r2 kQ (2 R ) 2 = kQ
4R 2 For C:
Since all of the charge is distributed evenly on the surface of the sphere, there is no
potential difference setup inside of the sphere and the electric field is zero everywhere
inside the sphere.
For D:
V= kq kQ
=
r
R For E:
The potential is the same everywhere inside the sphere, so
V= kQ
R Question 10:
Solution:
Using Gauss’s law, that the electric field due to the charge distribution is On the other hand, the electric potential at P1 inside the sphere is given by Solutions will be posted at www.prep101.com/solutions 20 ©Prep101 Phys1026 April Exam Solutions A plot of electric potential as a function of r is given as follows. 5.11 Solution Strategy – Conservation of Energy; EK (or KE) is converted to EP (or PE)
Each particle has KE = ½ mv2 = ½ (6.6 x 10–27 kg)(3.2 x 10–19 C)
∴ KEtotal = 2 [½ (6.6 x 10–27 kg)(3.2 x 10–19 C)]
KEtotal = 5.94 x 1014 J
At minimum separation, particles are motionless; KEtotal = 0 (ie: all energy is
converted to potential; in this case electric potential energy, or EPE)
∴ EPE = 5.94 x 1014 J (as stored between a pair of charges)
recall EPE = qV, and V = kq/r (for a point charge)
∴ 5.94 x 1014 J = kqq/r  solving for r to find min separation;
14 r = kqq/5.94 x 10 J r = (8.99 x 109Nm2/C2)(3.2 x 1019C)(3.2 x 1019C) / (5.94 x 1014 J)
r = 1.6 x 1014m
5.12 Answer d)
When the metal spheres are brought together, charge flows between them – since they
were equally and oppositely charged, the final charge on each is now 0 C. So the
electrical potential energy between them at any point is now zero.
When the insulated spheres are touched, only a little of the charge could flow – the
charge on the region where they touched. So the charge on each sphere is reduced. Solutions will be posted at www.prep101.com/solutions 21 ©Prep101 Phys1026 April Exam Solutions The potential energy of a pair of charged spheres is U = k q1q2/r.
Doubling the distance will half the potential energy, down to 5 J. But we reduced the
charges, so the sizes of the charges is lower, and so U is less negative. For example, 10
x (10) is 100, but 9 x (9) is 81. So, reducing charges relatively increased t...
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 Winter '14
 Prof.PaulineBarmby
 Physics

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