Problem 1
→
Sebuah benda bermassa 20 gram dan bermuatan q = +0,5 μC digantungkan pada seutas tali ringan r ingan yang massanya dapat diabaikan. Tepat disebelah kanan
benda pada jarak 15 cm diletakkan muatan q' = −1 μC yang menyebabkan menyebabkan 1
m A. 0,20 B. 0,24 C. 0,28 D. 0,32 E. 0,40 (Soal UMPTN 1995 Fisika)
9
2
2
posisi benda menjadi seperti gambar. Jika / 4πεo 4πεo = 9 x 10 Nm /C dan g = 10 −2 s , tegangan pada tali dekat pada harga .....Newton
Solution
We have three forces that are acting on q1, weight of q 1, rope tension and Coulomb force. Find the weigth of q1 first: −3 −1 W = mg = 20 ⋅ 10 (10) = 2⋅ 10 Newton Coulomb force from interaction with q 2 kq q 2 Fc = 1 2 / r (9 109)(0.5 ⋅10−6)(10−6) −2 2 Fc = ⋅ / (15 (15⋅10 ) Fc = 0.2 Newton
T = √[(Fc)2 + (W)2] Insert your data and we get T = 0.28 N Take a note that T , F c and W are in equilibrium condition, so then
T = F c + W (in
vectors sum symbol)
Proton yang bergerak dari keping A ke B seperti terlihat pada gambar, memperoleh kecepatan kecepatan 2 x −1 −27 5 10 m s . Jika antara dua keping vakum, d = 1 cm, dan massa proton = 1,6 x 10 kg, muatan 19 proton = 1,6 x 10 C, maka beda potensial keping sejajar tersebut adalah...volt (Soal UMPTN 1994 Fisika) Solution
Using mechanic energy conservation law:
Problem 5 An
→
5
electron travels into two plates with speed of 10 m/s. The electric field 4 between two plates is 10 N/C. Determine the value of d when the electron reaches the positive plate!
Diberikan dua buah keping sejajar seperti gambar. Sebuah elektron 5 bergerak dengan kecepatan 10 m/s memasuki keping. Kuat medan 4 listrik antara kedua keping adalah 10 N/C, tentukan nilai jarak d saat
elektron mencapai keping positif! (Fisika Study Center) Solution
Consider the case as projectile motion, the horizontal distance d
Find the acceleration of electron in the plates caused of coulomb force
The vertical distance of electron to get the time for reaching the positive plate
Back to horizontal distance and get d
Problem 6
→ Titik A terletak dalam medan listrik. Kuat medan listrik di titik A= 0,5 N C. Jika di titik A diletakkan benda bermuatan listrik 0,25 C, maka pada benda tersebut bekerja gaya Coulomb sebesar.... A. 0,125 N B. 0,25 N C. 0,35 N D. 0,40 N E. 0,70 N (Soal EBTANAS Fisika Tahun 2000)
Solution –1
Electric field E = 0.5 NC
Electric charge q = 0.25 C F = qE F = (0.25)(0.5) = 0.125 N
Problem 7
What must be the distance between point charge q 1 = 25.0 μC and point charge q2 = −64.0 μC for the electrostatic force between them to have a magnitude of 3 .60 N? → Tentukan jarak antara dua buah muatan titik q 1 = 25,0 μC dan q 2 = −64,0 μC agar gaya listrik statis antara kedua muatan sebesar 3,60 N ! Solution
Using the basic of electrical force formula
Problem 8
Given 4 particles form a square of edge length d = 10 cm with charges q 1
= −10 μC, q 2 = +20 μC, q 3 = −20 μC, and q 4 = +10 μC. Find the net electric field produced by the particles at the square's center!
→ 4 buah muatan membentuk suatu persegi dengan panjang sisi d = 10 cm. Keempat muatan masing-masing q 1 = −10 μC, q 2 = +20 μC, q 3 = −20 μC, and q4 = +10 μC.Tentukan besar kuat medan listrik yang dihasilkan muatan-muatan tersebut pada pusat persegi!
Solution
These are the directions of electric fields that are produced by 4 charges. To simplify the calculations, find the sum E 1 and E2, first find the distance of each charge to the center of square.
The distances are 0.5d√2 E1 and E3
E13 direction is the same with E 3 direction.
E2 and E4
E24 in the same direction with E 2 Finally the sum of E13 and E24 , E13 = E24 = x
o
(E13 and E24 forms an angle of 90 ) 9 −6 −1 Proceed with the data, k = 9 x 10 , μ = 10 , r = 0.5d√2 and d = 10 m.
Question 1 Two particles with electric charges are R meter away. The presence electric force is F. If the distance between two charges is changed to be 4R, find the ratio of electric forces presence in 4R and R distances! Answer
Electric force formula
so then
Question 3
Given two arranged charges Q 1 and Q2 as shown in the figure! 9
2
−2
Q1 = + 1 μC, Q 2 = − 2 μC and k = 9 x 10 N m C , then find the electric field of point P and its diection! Point P is located between two charges 4 cm away from Q1 Solution
Electric field formula resulted by a charge at a point that i s r away from the charge
We have two different E at point P
Use above formula and then get the sum of two electric field produced
The direction of resultant is to the right.
Question 4
Given three charges A, B and C form a right triangle shown our figure below. Find the resultant of electric force at charge A if F AB = FAC = 5 Newton! Solution
The angle formed by F AB and FAC 90° so using resultant formula then we get
Question 5
Three charges form an isoceles triangle as shown below, separated 10 cm away each others. 9
2
−2
Q1 = + 1 C, Q2= Q3 = − 2 C and k = 9 x 10 N m C force resultant at charge Q1 !
find the electric
Answer
As our previous question with the angle of 60° , and we get
Keep 18 x 10
11
N as X for easier calculation
Problem 6
Two charges that are Q 1 = 1 μC dan Q 2 = 4 μC 10 cm in distance. Find the location of a point where its magnitude of electric field is 0 N/C!
Pembahasan
There are three locations of possibilities, at the left of Q 1, at the right of Q2 or betwen the two.
The most possible locations that will produce zero electric field at point P are at t he left of Q1 and the at the right of Q 2. Try to take the first posiblity, consider the point P distance as x.
Problem 7 A negative electric charge of Q Coulomb rest in a homogenous electric field E with its direction is to the south. Determine the force acting on the charge and its direction! Answer
The relationship of electric field E, electric force F and charge: F = QE
where:
F and E in the same of direction for a positive electric charge F and E in the opposite of direction for an negative charge
By applying two statement above we have the direction of electric charge is to the North, cause of its negative charge.
Problem 8
Three charges around point P are shown in below figure!
9
2
−2
−12
−12
Use k = 9 x 10 N m C , Q1 = + 10 C, Q2 = + 2 x 10 −12 C dan Q3 = - 10 C, find the electric potential of point P ! Solution
The sum of electric potential at point P could be expressed as
Problem 9 8 charges form a cube as shown our figure in below. 4 charges with
+ 5 C of electric charge and the rest are
− 5 C . The cube side lenght is
r.
Find the magnitude of electric potential at point P, which is the center mass of the cube !
Solution
Applying the sum of electric potential at a point cause of 8 charges
Problem 10
Two charged particles at distance x tied by massless strings of lengts L shown below! Use tan θ = 0,75 and the strings tension are 0,01 N, find the electric force that is acting on each charge! Answer
Projection of forces acting on Q2 Apply equilibrium conditions for this case: FC = T sin Θ FC = 0,01 x 0,6 = 0,006 Newton
Problem 11
An negative charge of 5 Coulomb rests between two charged plates A and B. The electric force acts on the charge is 0,4 N with its direction is to the plate B as shown in our figure above, find the magnitude of electric field of plates A B and the sign of plate A! Solution F = QE E = F / Q = 0,4 / 5 = 0,08 N/C
and the sign of plate A is negative.
Problem 12
A ball conductor with electric charge of Q and its radius isi 10 cm as shown in the figure below. If the electric potential at point P is (kQ / x ) volt, find the
magnitude of x ! Solution
Use this equation to find the magnitude of electric potential at a point outside of the ball, V = (kq)/r where r is the distance of the point from the center of the ball or x = (0,1 + 0,2) = 0,3 meter.
Problem 13
Determine the work needed to move a positive charge of 10 μC from 230 kilovolt of electric potential to 330 kilovolt ! Solution W = q ΔV W = 10μC x 100 kvolt = 1 joule
Problem 14 Given a bal conductor l as shown below figure! E is the electric field at any points at certain distance from the center of the ball that its charge is +q.
Find the elecric field at points P, Q and R use x as the ball radius and point R at distance h from the ball surface! Answer
point P is inside the ball so then EP = 0 2 point Q is in the surface of the ball so then EQ = (kq)/x 2 point R is outside of the ball so then ER = (kq)/(x + h)
Problem 15
A particle of mass m with negative charge rests betweeb two paralell plate that are different in sign. Use g as the acceleration due to gravity and Q as particle charge and then find the magnitude of electric charge between two plates and the sign of plate Q ! Solution
In this case we have an equlilibrium state at particle cause of its weight and electrostatics force at the other side, consider these statement to find the sign of Q.
F and E in the same of direction for a positive electric charge F and E in the opposite of direction for an negative charge
Then we have the sign of plate Q is negative, and the magnitude of of electric field E F coulomb = W qE = mg E = (mg)/q
prep: fisikastudycenter.com