Substance:
- Determining the magnitude of the effects of electric charge such as Coulomb Force, Electric Field Strength.
- Determining the Position of the Zero Point
- Distinguishing between scalar quantities and vector quantities in terms of usage
Question No. 1
Two charged particles are at a distance R from each other and an attractive force of F occurs. If the distance between the two charges is made 4 R, determine the comparative value of the attractive force that occurs between the two particles to their initial conditions!
Solution 1
Question No. 2
Three charges A, B and C are arranged as shown below!
If QA = + 1 μC, QB = - 2 μC, QC = + 4 μC and k = 9 x 109 N m2 C- 2 determine the magnitude and direction of the Coulomb force on charge B!
Solution 2
On charge B, 2 forces work, namely the result of the interaction between charges A and B, called FBA, which is directed to the left and the result of the interaction between charges B and C, called FBC, which is directed to the right. Illustration as in the following picture:
Because the two forces are in a straight line but in opposite directions, to find the resultant force, simply subtract the two forces, for example, give the resultant a name
Total:
- F total = FBC - FBA
- F total = 72 X 10 - 3 - 18 x 10 -3 = 54 x 10 -3 N
- The direction according to FBC is to the right.
Question No. 3
Two charges are arranged as shown in the following image!
If Q1 = + 1 μC, Q2 = - 2 μC and k = 9 x 109 N m2 C- 2 determine the magnitude and direction of the electric field strength at point P which is located 4 cm to the right of Q1!
Solution 3
The basic formula used for this question is
Where E is the electric field strength produced by a charge, and r is the distance of the point from the source charge. Please remember again to determine the direction of E: "out of the positive charge" and "into the negative charge". Pay attention to the illustration in the picture!
The next step is to calculate each magnitude of the magnetic field strength E1 and E2 then find the resultant, don't forget to change the centimeter unit to meter. To make it easier, calculate separately one by one.
Direction to the right.
Question No. 4
The following image is an arrangement of three charges A, B and C which form a triangle with a right angle at A.
If the force of attraction between charges A and B is equal to the force of attraction between charges A and C of 5 F each, determine the resultant force on charge A!
Solution 4
Since the two forces form a 90° angle, find it using the usual vector formula:
Question No. 5
Three charges form an equilateral triangle as shown in the following figure. The distance between each of the three charges is 10 cm.
If Q1 = + 1 C, Q2= Q3 = - 2 C and k = 9 x 109 N m2 C- 2 determine the magnitude of the resultant Coulomb force on charge Q1!
Solution 5
The type of question is similar to question number 4, with an angle of 60° and the value of each force must be found first.
The number 18 x 1011 N, just call it X to make further calculations easier.
Question No. 6
Two charges, Q1 = 1 μC and Q2 = 4 μC, are separated by a distance of 10 cm.
Determine the location of the point that has zero electric field strength!
Solution 6
The location of the point is unknown so there are three possibilities, namely to the left of Q1, to the right of Q2 or between Q1 and Q2. To choose the correct position, pay attention to the following illustration and remember that the electric field strength "exits for positive charges" and "enters for negative charges". Just name the point to be searched as point P.
There are 2 places where E1 and E2 are opposite to each other, just take the point which is closer to the charge with smaller absolute value which is on the left of Q1 and call its distance as x.
Question No. 7
A negative electric charge of Q is in an electric field E that is directed to the south. Determine the magnitude and direction of the electric force on the charge!
Solution 7
The relationship between the electric field strength E and the electric force F that occurs on a charge q is
F = QEWith the following sign agreement:
- For positive charge, F is in the same direction as E.
- For negative charge, F is opposite to the direction of E.
In the question above, E is directed to the south so the direction of F is to the north, because the charge is negative.
Question No. 8
Look at the image of three charges located around point P below!
If k = 9 x 109 N m2 C- 2 , Q1 = + 10-12 C, Q2 = + 2 x 10-12 C and Q3 = - 10-12 C, determine the magnitude of the electric potential at point P!
Solution 8
Question No. 9
8 electric charges, 4 of which are + 5 C and the other 4 are - 5 C, are arranged to form a cube with a side length of r.
Determine the magnitude of the electric potential at point P which is the center of gravity of the cube!
Solution 9
Why zero? The distance of each charge to point P is the same and the magnitude of the charge is also the same, half positive and half negative so that if the numbers are entered the result is zero.
Question No. 10
Two particles with the same charge are suspended by a string so that they are arranged as shown in the following picture!
If tan θ = 0.75 and the tension in each rope is 0.01 N, determine the magnitude of the repulsive force between the two particles!
Solution 10
Pay attention to the style description in Q2 below!
Because the value of the rope force is known, then using the principle of normal balance we get:
- FC = T sin Θ
- FC = 0.01 x 0.6 = 0.006 Newtons
Question No. 11
A particle with a negative charge of 5 Coulombs is placed between two plates with opposite charges.
If the charge experiences a force of 0.4 N towards plate B, determine the magnitude of the electric field strength and the type of charge on plate A!
Solution 11
- F = QE
- E = F / Q = 0.4 / 5 = 0.08 N/C
For a negative charge, the direction of E is opposite to F, so that E is directed to the left and thus plate B is positive, plate A is negative.
Question No. 12
A hollow sphere has a charge of Q Coulomb and a radius of 10 cm.
Solution 12
To find the potential of a point outside the sphere, V = (kq)/r where r is the distance from the point to the center of the sphere or x = (0.1 + 0.2) = 0.3 meters.
Question No. 13
Determine the amount of work to move a positive charge of 10 μC from a potential difference of 230 kilovolts to 330 kilovolts!
Solution 13
W = q ΔV
W = 10μC x 100 kvolt = 1 joule
Question No. 14
Look at the following image! E is the electric field strength at a point caused by a hollow ball with an electric charge of + q.
Determine the magnitude of the electric field strength at points P, Q and R if the radius of the sphere is x and point R is a distance h from the surface of the sphere!
Solution 14
- Point P is inside the sphere so that EP = 0
- Point Q on the surface of the sphere so that EQ = (kq)/x2
- Point R is outside the sphere so ER = (kq)/(x + h)2
Question No. 15
A particle with mass m and negative charge is suspended between two parallel plates with opposite charges.
If g is the acceleration due to gravity and Q is the particle charge, determine the value of the electric field strength E between the two plates and the type of charge on the plate Q!
Solution 15
If we look at the forces acting on the particle, there is a gravitational force/weight force that is directed downward. Because the particle is floating, which means there is a balance of forces, then the direction of the electric force must be upward to balance the weight force. A negative charge means that the direction of the electric field E is opposite to the direction of the electric force F so that the direction of E is downward and the P plate is positive (E "out of positive, into negative"), the Q plate is negative.
To find the magnitude of E:
- F electricity = W
- qE = mg
- E = (mg)/q
Question No. 16
An electron with a mass of 9.11 × 10-31 kg and an electric charge of - 1.6 × 10-19 C, is released from the cathode towards the anode which is 2 cm away. If the initial speed of the electron is 0 and the potential difference between the anode and cathode is 200 V, then the electron will arrive at the anode with a speed of..
A. 2.3 × 105 m/s
B. 8.4 × 106 m/s
C. 2.3 × 107 m/s
D. 3 × 107 m/s
E. 2.4 × 108 m/s
Solution 16
Data from the question:
me = 9.11 × 10-31 kg
Qe = - 1.6 × 10-19 C
ν1 = 0 m/s
ΔV = 200 volts
ν2 = ....... !?
By the law of conservation of mechanical energy, the mechanical energy of electrons at the anode is the same as the mechanical energy at the cathode:
The origin of the formula is here,
This question is in non-calculator mode, you are not allowed to use a calculator to solve it, the alternative calculation is like this:
