Problem #1
The q charge is as large but different as far as a. The x point is located in the middle of the two charges. Electric field strength at point x of 180 N/C. How strong is the field at point x if point x is shifted a/4 towards one of the charges.Answer;
The electric field strength at the X point due to the two charges is
180 = kq/(a/2)2 + kq/(a/2)2 = 8kq/a2
kq/a2 = 22,5 N/C
If point x is shifted a/4 to one charge (see picture (ii)) then the strength of the electric field is
EX’ = kq/(a/4)2 + kq/(3a/4)2
= 16kq/a2 + 16kq/9a2
EX’ = 153/9(kq/a2) = (153/9)(22,5)
Ex’ = 382,5 N/C
Problem #2
A drop of oil m = 3.24 x 1011 kg falls vertically at a fixed speed between two charged parallel pieces. The electric field strength between the two pieces is 1.96 x 109 N / C. Determine the amount and type of oil drop load.
Oil drop mass m = 3.24 x 1011 kg; gravitational acceleration g = 9.8 m/s2, electric field strength E = 1.96 x 109 N/C with the downward direction, from the positive chip to the negative chip. If there is no force between pieces other than mg gravity, then the oil drops will move accelerated downward. But the fact is that oil drops move at a steady pace, meaning there is a force other than gravity which is the opposite of gravity. Because the oil drops are in an electric field, the force acting on the oil drops must have an electric force whose direction must be upwards which means the drops of oil must be negatively charged. With the magnitude of the electric force is F = Qe, then by applying Newton's I law,
ΣF = 0
+Fq – mg = 0
qE = mg, maka
q = (3,24 x 1011 kg)( 9,8 m/s2)/(1,96 x 109 N/C)
= 1,62 x 10-19 C
Problem #3
Two electric charges were 5.0 μC and –5.0 μC apart at a distance of 20 cm. (a) What is the strength of the electric field at point P which lies in the connection line of the two charges and is 5 cm apart from the positive charge? (b) How much electrical force is experienced by a charge of 3.0 μC if placed at point P?
Answer;
(a) Suppose q1 = 5.0 μC = 5 x 10-6 C and q2 = - 5.0 μC = - 5 x 10-6 C, then the location of point P is shown in the figure,
Electric field strength by q1, which is E1, directed to the right and magnitude,
E1 = kq1/r12 = (9 x 109 Nm2/C2)(5 x 10-6 C)/(0.05 m)2
= 1.8 x 107 N/C
The field strength is q2, which is E2 directed to the right, and the magnitude is
E2 = kq2/r22 = (9 x 109 Nm2/C2)(5 x 10-6 C)/(0.15 m)2
= 2 x 106 N/C
Because E1 and E2 are in the same direction, the resultant electric field strength is at point P, which is Ep directed to the right and magnitude
EP = E1 + E2 = 1.8 x 107 N/C + 2 x 106 N/C = 2 x 107 N/C
(b) the magnitude of the electric force experienced by a charge of 3.0 μC if placed at point P is
F = qE = (3 x 10-6 C)(2 x 107 N/C) = 60 N
Problem #4
Two loads of 32µC and -2.0 μC are separated at a distance of x. A test load placed at point P then released, it turns out that the charge remains silent. How big is x?
Answer;
Each charge q1 = 32μC and q2 = -2Μc, the test load at point P is not moving, this indicates that the number of electric fields due to the two charges at point P is zero. Then apply,
E1 = E2
q1/r12 = q2/r22
with r2 = 10 cm and r1 = (x + 10) cm than,
32μC/(x + 10)2 = 2 μC/(10)2
16 = [(x + 10)/x]2
(x + 10)/x = 4
x + 10 = 40
x = 30 cm
So the separation distance from the two loads is approximately 30 cm.
Problem #5
A particle charged 3.2 x 10-19 C and a mass of 6.4 x 10-17 kg was placed from rest in a homogeneous electric field of 4.0 x 104 N/C. If the electric field generated by the particle charge itself is ignored, determine (a) the speed of the particle at t = 0.02 seconds, and (b) the distance the particle has traveled for 0.05 seconds.
Answer;
Load particles q = 3.2 x 10-19 C, mass of particles m = 6.4 x 10-17 kg, initial velocity v0 = 0 m/s, electric field strength E = 4.0 x 104 N/C.
(a) the speed of the particle at t = 0.02 seconds,
The acceleration experienced by particles due to the electric force,
+ Fq = qE
qE = ma
a = qE/m, then
a = (3.2 x 10-19 C)(4.0 x 104 N/C)/(6.4 x 10-17 kg)
= 2.0 x 102 m/s2
The speed of particles at t = 0.02 seconds is
v = v0 + at
v = qEt/m
= (3.2 x 10-19 C)(4.0 x 104 N/C)(0.02 s)/(6.4 x 10-17 kg)
v = 4.0 x 1010 m / s
(b) the distance that the particle has traveled for 0.05 second is
s = v0t + ½at2
s = qEt2/2m
= (3.2 x 10-19 C)(4.0 x 104 N/C)(0.02 s)2/2(6.4 x 10-17 kg)
s = 2.5 x 109 m
Problem #6
At points A and C, an ABCD square whose sides are 3 cm each is placed with a particle charged +10 µC. In order for the electric field strength at B to be zero, determine the type and the amount of the point D load!
Answer;
Because the charge at A and at C and the distance AB and BC is equal, the electric field at point B due to the charge at point A (E1) is equal to the electric field at point B due to the charge at point C (E2), namely
E1 = E2 = E = kq/a2 = 9 x 109 x 10 x 10-6/(0,03)2 = 108 N/C
Then the big resultant is both
E12 = (E2 + E2)1/2 = E√2
In order for the total electric field at point A to be zero then it must be at point D a negative charge is placed so that an electric field arises at point B due to the charge at point D (E3) that is parallel to E12 but the same size, with the magnitude E3 is
E3 = kqD/rBD2 = E√2
Remember that rBD = a√2 = 0.03√2 m, then
9 x 109qD/(0,03√2)2 = 108√2
qD = 2√2 x 10-5 C = 0.2√2 μC
Problem #7
A small charged ball with a mass of 2 grams is hung on a light rope (assuming no mass) and is in the electric field area E = (3i + 4j) x 105 N/C. If the ball is in equilibrium at θ = 530. Determine (a) the size of the ball load and (b) the amount of tension of the rope!
Answer;
Known: E = (3i + 4j) x 105 N/C, Ex = 3 x 105 N/C, Ey = 4 x 105 N/C, m = 2 g = 2 x 10-3 kg;
The relationship between the electric field and the electric force is given by
F = qE or Fx = qEx and Fy = qE
Fqx = (3 x 105)q and Fqy = (4 x 105)q.
Note the forces acting on the x axis,
T sin 530 = Fqx = (3 x 105)q (1)
Note the forces acting on the y axis,
T cos 530 + Fqy = mg = (2 x 10-3)(10) = 0.02 or
T cos 530 = 0.02 - (4 x 105)q (2)
equations (1) and (2) are divided we get,
tan 530 = (3 x 105)q/[0,02 - (4 x 105)q]
4/3 = (3 x 105)q/[0,02 - (4 x 105)q]
25 x 105q = 80 x 10-3
q = 3.2 x 10-8 C
(b) rope tension can be obtained by using equation (1) or (2), i.e.
T sin 530 = Fqx = (3 x 105)q
T sin 530 = (3 x 105)(3.2 x 10-8 C)
T = 0.012 N
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