CH21-26.Problems
Exam 2 Review
JH
b and c tie, then a (zero)
The net force is zero at the origin and zero at the infinity and thus it must have
some maximum somewhere in between. The net force is given by:
To find the maximum, you must derive this with respect to x then make it equal
to zero. ( too long for an exam). The answer is when x = d/Sqrt(2)
Using the E field as force is not a good idea
The best is to convert the potential energy to kinetic energy:
e(V(inf) –V(R)) = -(Kf- Ki )
eV = ½ m v^2  v = Sqrt(2eV/m) = Sqrt(4 e kq/d /m) = 9.4e5 m/s
50. Figure 25-49 shows a parallel-plate capacitor of plate area A= 10.5 cm2 and plate separation
2d = 7.12 mm. The left half of the gap is filled with material of dielectric constant k1 = 21.0; the
top of the right half is filled with material of dielectric constant k2 = 42.0; the bottom of the
right half is filled with material of dielectric constant k3 = 58.0.What is the capacitance?
Answer: 4.55×10-11 = 45.5 pF