Solutions to Practice Problems for Exam 1
Coulomb force.
Calculate the electric force between an alpha particle (charge of 3.2x10-19 C) and an
aluminum nucleus (charge =20.8x10-19 C) located at a distance of 1.2x10-10m.
F = k Q1 Q2 / r2 = 9 x 109 N m2 / C2 (3.2x10-19 C) (20.8x10-19 C) / (1.2x10-10m) 2
= 4.2 x 10-7 N
Electric field.
Determine the electric field at the origin due to the two charges A and B. Consider the
positions given in meters and charge of A=1.1C and B=-1.5C.
EB
EA
EA = kQA / r2 = 9 x 109 N m2 / C2 (1.1x10-6 C) / (1 m) 2 = 9900 N/C
EB = kQB / r2 = 9 x 109 N m2 / C2 (1.5x10-6 C) / (1 m) 2 = 13500 N/C
ET = E A2  E B2 = 16740 N/C
θ = tan-1(EA/ EB) = -36.3o
Electric potential.
A 35C charge is placed 0.65m from an identical 35C charge. Calculate the electric
potential at the point midway between them.
V1 = k Q1 / r = 9 x 109 N m2 / C2 (35x10-6 C) / 0.325 m = 969000 V
V2 = V1 = 969000 V
VT = V1 + V2 = 1.94 MV
Capacitance, dielectrics:
Calculate the capacitance of the two plates shown in the figure when they are completely
submerged in a liquid whose dielectric constant, K is 3.4. Area of the plates=0.24 m 2,
distance between the plates = 0.36 mm.
[Dielectric permittivity of vacuum,  o  8.85 x10 12 F / m ]
d = 0.36 mm = 3.6 x 10-4 m
A
C   o
= 3.4( 8.85 x10 12 F / m )(0.24 m2) / (3.6 x 10-4 m) = 2.0 x 10-8 F or 20 pF
d
Capacitance, Dielectrics and Storage of Electric Energy.
A cardiac defibrillator is used to shock a heart that is beating erratically. A capacitor in
this device is charged to 2.5 kV and stores 950 J of energy. What is the capacitance?
Energy = ½ C V2
C = 2 E / V2 = 2 ( 950 J) / (2500 V) 2 = 3.0 x 10-4 F or 300 µF