Problem #1
What is the electric force acting on electrons by protons (hydrogen atom nuclei) when electrons surround the proton at an average distance of 0.53 x 10-10 m (Bohr jejari) with k = 9 x 109 Nm2/C2 and e = 1.6 x 10-19 C.Answer;
Proton charge q1 = +1.6 x 10-19 C, electron charge q2 = - 1.6 x 10-19 C, distance of electron proton r = 0.53 x 10-10 m. The electrical force on an electron by a proton is. . . .?
F = kq1q2/r2
= 9 x 109 Nm2/C2 (1,6 x 10-19 C)(1,6 x 10-19 C)/(0,53 x 10-10 m)2
= 8,2 x 10-8 N
Problem #2
Suppose that two electric charges separated by 10 cm experience pull-pull force 12 N. What is the attractive force between the two if they are 5 cm apart?
Answer;
Separation distance r1 = 10 cm, F1 = 12 N, r2 = 5 cm, Force F2 ?
From the Coulomb force equation F ~ 1/r2 so
F1/F2 = (r1/r2)2
F2 = (r1/r2)2 F1
= (10 cm/5 cm)2 x 12 N = 48 N
Problem #3
Two electric charges +5 x 10-5C and +9.8 x 10-5C separate at a distance of 7 cm. Determine (a) the Coulomb force if the two charges are in air, (b) the Coulomb force when the two charges are in a material whose permeativity is ε = 1.77 x 10-5 C2/Nm2, (c) the relative permittivity of the material is a matter of parts ( b).
Answer:
Each charge is q1 = +5 x 10-5C and q2 = +9.8 x 10-5C, r = 7 cm = 7 x 10-2 m.
(a) Coulomb force if the two loads are in the air
F = kq1q2/r2
= 9 x 109 Nm2/C2(+5 x 10-5C)(+9.8 x 10-5C)/(7 x 10-2 m)2
= 9 kN
(b) The size of the Coulomb force if the two muata are in a material
F = (1/4πε)(q1q2/r2)
= [1/4π (1.77 x 10-5 Nm2/C2)] (+ 5 x 10-5C)(+9.8 x 10-5C)/(7 x 10-2 m)2
= 4.5 kN
(c) Relative permeativity is calculated by ε = εrε0
εr = ε/ε0 = (1.77 x 10-5 Nm2/C2)(8.85 x 10-12 C2/Nm2) = 2.00
Problem #4
Two electric charges each - 4μC and 10μC separate at a distance of 12 cm. Between the two loads is placed a third charge at a distance of 3 cm from the + 10μC charge. (a) Determine the magnitude and direction of the force on the third charge which is +5 μC ?, (b) Determine the magnitude and direction of force on the charge when the magnitude is -5µC?
Answer:
(a) The third position of the charge is described as follows
Each charge, q1 = 4.0 x 10-6 C, q2 = 10 x 10-6C, q3 = 5 x 10-6C and r13 = 9 x 10-2 m and r32 = 3 x 10-2 m.
The set of horizontal directional force vectors is positive. The electric force experienced by q3 is due to the pull of q1, which is F31 to the left, so it is negative.
F31 = - kq3q1/r312
= - 9 x 109 Nm2/C2 (+4 x 10-5C)(+5 x 10-5C)/(9x 10-2 m)2
= - 200/9 N
The electric force experienced by q3 is due to q2 repulsion, which is F32 to the left, so it is negative.
F32 = - kq3q2 / r322
= - 9 x 109 Nm2/C2 (+5 x 10-5C)(+10 x 10-5C)/(3 x 10-2 m)2
= - 4500/9 N
So, the resultant force experienced by q3, namely F3 is
F3 = F31 + F32
= - 200/9 N + (- 4500/9 N)
F3 = - 522 N
A negative sign indicates F3 to the left.
Problem #5
Two small balls each charged 135 µC and –60 μC with 10 cm apart. Where should the third charge be placed so that the resultant force acting on the charge is zero?
Each charge, q1 = 13.5 x 10-5 C, q2 = 6.0 x 10-5C, and r12 = 0.1 m. If the third charge is placed in the space between q1 and q2, it will not produce a resultant force equal to zero. This is because the Coulomb force of q1 and q2 is always one-way.
So, the third charge must be placed outside the space but still in line, which is near the smallest charge so that the result can be zero. Therefore the third charge q3 must be placed near q2 shown in the image above. In order for the resultant force acting on q3 to be zero, it is clear that F31 must be equal to F32. For example r32 = x m, then r31 = 0.1 + x.
F31 = F32
kq3q1/r312 = kq3q2/r322
q1/q2 = r312/r322
135 μC/60 μC = [(0,1 + X)/X]2
(0,1 + X)/X = 3/2
0,1 + X = 1,5x or x = 0,2 m
Thus, the third charge must be placed 0.2 m from the load –60 μC or 0.3 m from the load of 135 μC
Problem #6
Two small balls like the following picture each have a mass of 2.0 g and have the same unit. One ball is hung with a true isolation. The other ball below approaches the ball that is hung so that the ball that is hung up and is stationary when the angle formed by the thread is vertical is 370 and the distance between the two balls is 3.0 cm. Determine (a) the weight of the ball, (b) the magnitude of the electric force, (c) the electric charge on each ball. (g = 9.8 m.s─2)
The mass of the ball m = 2.0 g = 0.002 kg, acceleration of gravity, g = 9.8 m/s2, the distance between the two balls r = 3 cm = 0.03 m.
(b) the component of the rope tension force is
Tx = T sin 370 = 0.6T and Ty = T cos 370 = 0.8T
The ball is silent and balanced on the x axis, then
Fq = Tx = 0.6T (1)
On the y axis
Mg = Ty = 0.8T (2)
Equation (1) and (2) are divided we get
Fq/mg = 0.6T/0.8T
Fq = ¾ mg
= ¾ (0.0196 N)
Fq = 0.0147 N
(c) Suppose the electric charge on each ball is q then the electric force;
F = kqq/r2 = kq2/r2
q2 = Fqr2/k
q = r [Fq/k]1/2 = 0.03 m x [(0.0147 N/9 x 109 Nm2/C2)]1/2
q = 3.83 x 10-9 C = 3.83 nC
Problem #7
The following diagram shows three small electrically charged balls.
The force acting on ball B is? (k = 9 x 109 Nm2/C2)
Answer:
Note the repelling force and pull between the charge B with the charges A and C below!
FBA = kqAqB/rAB2 = (9 x 109 Nm2/C2)(40 x 10-6 C)(10 x 10-6 C)/(0.2 m)2
FBA = 90 N
And
FBC = kqCqB/rCB2 = (9 x 109 Nm2/C2)(30 x 10-6 C)(10 x 10-6 C)/(0.2 m)2
FBA = 67.5 N
Then the force acting on the charge B ball is
FB = (FBA2 + FBC2)1/2
FB = [(90 N)2 + (67.5 N)2]1/2
FB = 112.5 N
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