Chapters 25 and 26
Electric Potential of a Point Charge:
The discussion of electric potential is important because we are always looking for convenient sources of energy. Since any two point charges exert a force of attraction or repulsion on each other, if one charge moves in the field of the other a distance dr under an average force F, the work done is equal to Fdr. This means that if there is just a single charge alone in the entire space, there is potential. When a second charge is placed in the field of the first charge, then work and energy actualize and we think of potential energy. Recall the "potential energy" concept you learned in Physics I. The cause of gravitational potential energy is gravity. The potential energy a rock has in the gravitational field of the Earth is because of gravity. Gravity attracts mass M with a force F = Mg. When a rock is lifted to a height h, the work done on it is equivalent to Mgh. The potential energy stored or consumed is also Mgh.
A similar concept applies to the electrostatic field that charge +q_{1} creates at a distance r_{1} from it: E = kq_{1}/r^{2}. If another charge +q_{2} is placed at r_{1}, q_{1} repels it. As q_{2} moves away under the varying and repelling force F=kq_{1}q_{2/}r^{2} to ∞, more and more work will be done on it. The maximum amount of work that can be done may be calculated by integrating F∙dr from r_{1} to ∞.

The electric potential V of
a point charge q_{1} at
a typical point P in
space at a distance r_{1} from
it is given by :

Now
if another charge like q_{2}
is placed at P a
distance r_{1} from q_{1} ,
then q_{2} finds
a potential energy equal to

Consequently, we can write : P.E. = V_{1 }q_{2} or, in general, U_{e} = Vq.
If we examine the unit of V, we will see that it has units of (energy per charge) or in SI units (Joules / Coulomb). Let's do this examination. Also let [ ] denote "the unit of ".
(Joule / Coul ) is called ( Volt ). 1 Volt means 1 J/Coul.
A charge in space generates different Potentials at different distances from it. The presence of a second charge is necessary for Potential Energy to make sense.
Example 1: Calculate the electric potential of q_{1} = 25.0nC at (a) 1.00m, (b) 2.00m, and (c) infinite long distance from it.
Solution: (a) V_{1} = kq_{1} / r_{1} = [(9.00x10^{9})(25.0x10^{9}) / (1.00)] J/Coul. = 225 Volts.
(b) V_{2} = kq_{1} / r_{2} = [ (9.00x10^{9})(25.0x10^{9}) / (2.00)] J/Coul. = 113 Volts.
(c) V_{3} = kq_{1} / r_{3} = [ (9.00x10^{9})(25.0x10^{9}) / ( ∞ )] J/Coul. = 0.
Example 2: Calculate the potential energy that another charge q_{2 }= 5.00nC possess when placed at the three different points of the previous example.
Solution: P.E._{1} = V_{1}(q_{2}) = 225(J/Coul.) * (5.00x10^{9} Coul.) = 1130 nJ.
This means that it takes 1130 nJ of energy to push a 5.00nC of positive charge from infinity to a distance of 1.00m from charge q1.
P.E._{2} = V_{2}(q_{2}) = 113(J/Coul.) * (5.00x10^{9} Coul.) = 565 nJ.
It takes 565 nJ to push a 5.00nC charge from infinity to a distance of 2.00m from q1.
P.E._{3} = V_{3}(q_{3}) = (0.0) * (5.00x10^{9} Coul.) = 0.0 nJ.
This means that it takes no effort (energy) to place a charge very far away from q_{1}.
Example 3: How much energy is needed to place 1.00μCof charge at each corner of an equilateral triangle 0.250m on each side? Suppose that each charge is coming from far away (infinity) and that the triangle itself is far away from other electric charges.
Solution: 1) Placing the 1st charge does not require any energy because other corners are empty. There is no repelling force against the first charge and it can be done effortlessly (W_{1} = 0). 2) To bring a 2nd charge from infinity and place it at 0.250m from the 1st charge some work must be done. The work to be done is equal to the change in P.E. of the 2nd charge in the field of the 1st charge. It is equal to:
W_{1} = 0. W_{2} = kq_{1}q_{2} / r = [ (9.00x109)(1.00x106)(1.00x106) / (0.250) ] J = 36.0 mJ. 3. The 3rd charge faces resistance from both the 1st charge and the 2nd charge. W_{3} = { kq_{1}q_{3} / (0.250) + kq_{2}q_{3} / (0.250) }mJ = 36.0mJ + 36.0 mJ = 72.0 mJ. Finally, W_{total} = W_{1} +W_{2} +W_{3} = {0 + 36.0 +72.0} mJ = +108.0 mJ.

Fig. 1 
Potential in a Constant Electric Field:
Fig. 2 A constant electric field is the field in between two parallel sheets equally but oppositely charged. If the positive sheet is on the left and the negative on the right, as shown, the direction of E is to the right or along the (+x) axis. Since E is constant, so is F, the force on any point charge q placed in E. The Work done on +q by the constant F as it is pushed to the right varies linearly with x, simply W = F ∙ x. 
Since F = qE, the work done becomes: W = qE ∙ x = qEx (2) (Dot vanishes because Force and displacement are parallel). As the field does more work in pushing q to the right, more of its potential is used up. As the work done by the field (W) increases because of the increase in x, the P.E. or U_{e} of q decreases. We may write: W =  P.E. = U_{e}. Equation (2) becomes: U_{e} = qEx or, (U_{e} /q) = Ex. (3) The left side is energy per unit charge or potential, V. Equation (3) becomes: V = Ex. V is a linear function of x when E is constant. At x=0, potential V is maximum and equal to Ex_{1}. Why? At x = x_{1}, potential is zero. Why? This makes the V equation to be V = E(xx_{1}). 
Equipotentials:
Around a charge distribution, there are points at which potential is the same. This means that if charge q is placed at any of such points, it will have the same potential energy. A collection of such points form a surface that is called an "equipotential surface." In 2D, we show such surfaces as lines and call them "equipotential lines." For example, around a positive or a negative point charge, there are infinite concentric spheres that each is an equipotential surface. Of course, the each charge itself is at the center of such sphere. In the 2D drawings shown below, for each of the (+) or () charge, the concentric circles are in fact concentric spheres.
The above charges are of course assumed to be separate and very far from each other and any other charges.
Fig. 3
The important point is that anywhere a field line (red or blue) crosses an equipotential line (each black circle), the angle is 90 degrees; in other words, field lines are necessarily perpendicular the equipotential lines. In Lab, it is easy to find the equipotential lines for a certain charge distribution by using a voltmeter. Field lines can then be drawn keeping in mind that they must be at right angles to the equipotential lines. Below, the fieldlines and eqipotential lines are shown for an electric dipole and a parallelplates capacitor.
Fieldlines and Equipotential Lines of a Dipole Fig. 4 
Fieldlines and Equipotential Lines in between the plates of a parallelplates capacitor Fig. 5 
As long as a charge like q_{2} travels on one equipotential surface (caused by another charge like q_{1}), its potential energy remains constant. When q_{2} travels from one equipotential surface to another one, its energy changes. The change in energy is equal to the work done on the charge that can be positive or negative. The path that q_{2} takes in going from one equipotential surfcae to another, is not important. What is important is the potential difference between the two equipotentials. In the following figure, two of the infinite equipotentials around charge q_{1} are shown. The energy change that q_{2} experiences as it goes from equipotential 1 to equipotential 2 is the same regardless of the path taken.
Fig. 6
Test Yourself 1: click here.
1) The electric field strength E_{1} of a point charge q_{1} at a distance r is (a) E_{1} = kq_{1}/r^{2} (b) E_{1} = kq_{1}/r (c) E_{1} = kq_{1}/r^{3}.
2) The electric force F of field E_{1}on charge q_{2} is (a) F = E_{1}q_{2} (b) F = {kq_{1}/r^{2}}q_{2} (c) both a & b. click here.
3) The electric potential V_{1} of a point charge q_{1} at a distance r is (a) V_{1} = kq_{1}/r (b) V_{1} = kq_{1}/r^{2} (c) V_{1} = kq_{1}/r^{3}.
4) The potential energy P.E. of point charge q_{2 } at points in space where the potential is V_{1} is (a) P.E. =V_{1 }q_{2 } (b) P.E. =_{ }(kq_{1}/r)_{ }q_{2} (c) P.E. =_{ } kq_{1}q_{2 }/r (d) a, b, and c. click here.
5) The first 4 questions and their correct answers apply to point charges (a) True (b) False.
6) The forcefield formula F = Eq is true (a) if E is caused by a point charge (b) if E is uniform and caused by a parallelplates capacitor (c) both a and b.
7) The definition of the electric potential, V, at a point is (a) K.E. per unit charge at that point (b) Force per unit charge at that point (c) Electric P.E. per unit charge at that point. click here.
8) The definition of the electric field, E, at a point is (a) E = force per distance at that point (b) E = force per unit charge at that point (c) E = kinetic energy per unit charge at that point. click here.
9) The electric potential energy, P.E., is (a) P.E. = Vq (b) P.E. = Vq_{2}^{2} (c) P.E. = (K.E.)q_{2}.
10 ) The reason for using q_{2} in the above questions instead of just q is that (a) q_{2 }is the charge that is placed in the field of q_{1} (b) potential, V, field, E, and force, F, in the above formulas are caused by charge q_{1} (c) both a & b.
11) Even if we do not use q_{2 }instead of simply q, and write F = Eq and P.E. = Vq, it is understood that (a) E and V are caused by a charge other than q (b) q is placed in the field E of a different charge (c) both a & b.
12) The SI unit of E, the electric filed, is (a) C/m (b) N/C (c) N/m. click here.
13) The SI unit of V, the electric potential, is (a) C/s (b) Joules/C (c) Joules/m.
14) The potential ,V, at 9.0m from a +25μCoul. charge is (a) 2778 J/C (b) 25000 J/C (c) 0.
15) If +1.0Coul. of charge is placed at 9.0m from the charge in Question 14, it finds a potential energy equal to (a) 2778J (b) 0 (c) 25000J. click here.
16) If 1.0Coul. of charge is placed at 9.0m from the charge in Question 14, it finds an energy of (a) 2778J (b) 0 (c)  25000J.
17) The reason why the answer in Question 16 is negative is that (a) work has to be done to move the negative charge to infinity under the attraction of the positive charge (b) in moving the negative charge away, displacement is outward while the attraction force is inward (c) both a & b. click here.
18) The potential at 3.0m from a 15.0μC charge is (a) 45000J/C (b) 15000J/C (c) 45000J/C.
19) If 40.0μC of charge is placed at 3.0m from the charge in Question 18, it finds a potential energy of (a) 4.5J. (b) 1.8J. (c) 9.0J.
20) If 40.0μC of charge is placed at 3.0m from the charge in Question 18, it finds a potential energy of (a) 4.5J. (b) 1.8J. (c) 1.8J. click here.
21) The energy it takes to place a 4.0μC charge at a corner of an equilateral triangle (2.0m long on each side) that has no charge on it and is far from other charges is (a) zero (b) 2.0J (c) 2.0J.
22) If in Question 21, if one corner has that 4.0μC charge, the energy it takes to place another 4.0μC charge at a 2nd corner is (a) 0.144J (b) 0.072J (c) 0.144J. click here.
23) In Question 22, to place another 4.0μC charge at the 3rd corner, it takes (a) 0.144J (b) 0.072J (c) 0.072J
24) Formulas V_{1} = kq_{1}/r and P.E. =_{ } kq_{1}q_{2 }/r apply (a) to point charges only (b) surface charges only (c) both a & b.
25) In the space between a parallelplates capacitor, electric field, E is constant. Potential energy varies with distance from each plate. The way P.E. varies with (x), its distance from one of the plates, is (a) proportional to x (b) proportional to 1/x (c) proportional to x^{2}. click here.
Parallelplates Capacitor:
Two parallel and metallic plates separated by an insulator form a "parallelplates capacitor". Capacitors store electric energy. If two flat sheets of aluminum foil sandwich a thin sheet of paper, a parallelplates capacitor is formed. When aluminum foils are connected to the poles of a battery, electrons from the negative pole flow through the connecting wire and distribute themselves over one foil making it the negative plate. This negative plate (foil) repels equal number of electrons from the other plate (foil) and causes the other foil to become the positive plate. The repelled electrons flow toward the positive pole of the battery where they are wanted and get absorbed by it. The closer the plates (or the thinner the insulating material, here the paper), the more charge accumulation occurs on them. However, there is a limit to the amount of positive and equally negative charges that can accumulate themselves on the two plates (foils). If accumulation exceeds a certain amount, electric discharge takes place via a spark through the insulator. The internal spark will burn the insulator and the capacitor goes bad. We will come back to the discussion of parallelplates capacitors after the following general discussion on capacitors.
ChargetoVoltage Ratio for Capacitors:
Fig. 7 shows an initially empty capacitor that is connected to a battery. If key K is closed( turned on) at t = 0, the battery starts charging the capacitor. The voltmeter will show that the capacitor voltage V_{C} keeps increasing with the increase in charge (q). The reason is very obvious. As time (t) increases, the capacitor charge accumulation (q) increases that causes the electric P.E. of the capacitor to increase. As a result, the capacitor voltage, V_{C} increases. In brief, the more charge q on each plate, the greater the capacitor voltage V_{C}. Experiment verifies that the chargetovoltage ratio (q/V_{C}) or simply q/V for each capacitor is a constant and is called the "capacity C of the capacitor." We may write this as 
A voltmeter is placed across the capacitor to monitor its voltage, V_{C}. Resistance R controls the flow of charges to the capacitor and avoids sudden charging. R is like a valve in water systems that if opened slightly the flow will be controlled and small. Fig. 7 
The SI Unit of Capacity:
In SI, charge is in Coulomb and voltage in volt; therefore, capacity becomes Coul./volt called "Farad." The capacity of a capacitor is said to be 1 Farad (1F) if it can hold a maximum charge of 1C when connected to a voltage of 1V. Majority of capacitors have very small capacities. Most of them are built to hold charges of μC or nC amounts.
Example 4: Calculate the capacity of a capacitor that holds at most 30.0μC of charge when connected to a 12.0V battery.
Solution: C = (Q/V) = 30.0μC / 12.0volts = 2.50μC/volt or C = 2.50μF.
Example 5: When a capacitor is half charged, it has 60.0μC of charge on each plate and the voltage across it is 7.50 volts. Find its capacity.
Solution: The chargevoltage ratio is C = Q/V. The more the accumulated charges, the greater the voltage it makes. Capacity, C, is constant, anyway.
C = 60.0μC / 7.50 volts = 8.00μF (microFarads).
Back to Parallelplates Capacitor:
The capacity, C, of a parallelplates capacitor is directly proportional to the area of each plate (A) and inversely proportional to the insulator thickness (d). C is also proportional to a physical property of the insulating material shown by (ε) that is called the "permittivity." It is related to how well the insulating material allows the electric field lines to pass through. The Capacity, C, of a parallelplates capacitor is therefore given by
The quantity (ε) is called the Permittivity of the insulating material that is also called the "dielectric." The permittivity of vacuum (free space) is shown by ε_{o}. These two quantities are related by ε = κε_{o} where κ pronounced " kappa" is called the dielectric constant of the material. The value of κ for vacuum is 1, for mica is 5.4, and for water is 80. This means that if mica is used as the insulator, the capacity increases by a factor of 5.4 compared to vacuum or almost air.
Example 6: Calculate the capacity of a parallelplates capacitor with rectangular (20.0cm by 30.0cm) aluminum plates separated by a 0.10mm sheet of paper. The dielectric constant of regular paper is κ = 3.3.
Solution:

C =
3.3(8.85x10^{12} (F/m)
)* ( 0.200m X 0.300m) / ( 0.00010m)
= 17.5x10^{9} F
= 17.5
nF Note: 1 Farad of capacity is a very large capacity 
Example 7: Calculate the area of each plate of a 1.00Farad parallelplates capacitor with an air gap of 0.0500mm.
Solution: Solving the capacity formula for (A), yields: A = Cd /ε. Substituting yields:
A = (1.00 F)(0.0500x10^{3}m) / (8.85x10^{12} F/m) = 5.65x10^{6} m^{2}
This means a square of side 2380m = 1.48 miles (A capacitor 1.48mi by 1.48mi ? Not practical !!!)
Connection of Capacitors:
It is sometimes necessary to come up with a capacitor of a certain capacity that is not available. By combining two or more of different capacitors, the desired capacity can be made. In two ways capacitors may be connected: in series and in parallel. An equivalent capacity can be calculated for each type of connection. The following figure shows both types of connection and a formula that calculates the equivalent capacity for each type:
Fig. 9 Fig. 10
Series: The battery voltage must equal the sum of voltages across the three capacitors. We may write: V_{total } = V_{ab} + V_{bc} + V_{cd } (1) _{ }If 2 electrons flow to the left of C_{1}, they repel 2 electrons from the right plate of C_{1} making its right plate 2 units positive. Those repelled electrons move to the left side of C_{2} making it 2 while repelling 2 electrons from the right side of it making its right +2. The same happens to C_{3}. The repelled 2 electrons from the right of C_{3} will be absorbed by the positive pole of the battery and the flow for those 2 electrons is complete. Of course saying "2 electrons" is just an example. In reality some 10^{13} or 10^{14} more or less electrons might easily flow. Every capacitor in Fig. 9 (series) ends up with same amount of charge Q. Equivalent Capacity: The single capacitor that can replace those three capacitors must hold the same amount of charge, simply Q. For the equivalent capacitor, we may write: Q = C_{eq}V from which V = Q/C_{eq}. For each capacitor we write its q = CV. V_{ab} = Q/C_{1} ; V_{bc} = Q/C_{2} ; V_{cd} = Q/C_{3}. Substituting in (1), results in Q/C_{eq} = Q/C_{1} + Q/C_{2} + Q C_{3} Dividing through by Q, yields: 1/C_{eq} = 1/C_{1} + 1/C_{2} + 1/C_{3} Series Capacitors. 
Parallel: The total charge Q_{total} that leaves the battery distributes over the three capacitors such that Q_{total } = Q_{1 }+ Q_{2} + Q_{3}. Electrons arriving at e divide into 3 branches. (Fig. 10) If capacities C_{1}, C_{2}, and C_{3} are proportional to numbers 2, 3, and 4, for example, and say 18 electrons leave the negative pole of the battery, 4 will flow to C_{1}, 6 will flow to C_{2}, and 8 will flow to C_{3} and settle on their left plates. Equal number of electrons will be repelled from the right plates making them positive. The repelled 18 electrons will be absorbed by the positive pole of the battery and the flow for those 18 electron is complete. This is just an example, in reality some 10^{13} or 10^{14} more or less electrons could easily flow. Q_{total } = Q_{1 }+ Q_{2} + Q_{3}. Equivalent Capacity: Using Q = C_{eq}V for the equivalent capacitor as well as the individual capacitors, yields: C_{eq}V = C_{1}V + C_{2}V + C_{3}V. Dividing through by V, yeilds: C_{eq} = C_{1} + C_{2} + C_{3} Parallel Capacitors. 
Look at the following two simple examples:
Example
8: A
30.0μF capacitor is in series
with a 6.00μF capacitor.
Find the equivalent capacity. Solution: 1/C_{eq} = 1/C_{1} + 1/C_{2} ; 1/ C_{eq} = 1/30.0 + 1/6.00 ; C_{eq} = 5.00μF. Make sure you use horizontal fraction bars when verifying the solution.

Fig. 11 
Example
9: A
30.0μF capacitor is in
parallel with a 6.00μC
capacitor. Find the equivalent capacity. Solution: C_{eq} = C_{1} + C_{2} ; C_{eq} = 30.0 + 6.00 ; C_{eq} = 36.00μF.

Fig. 12 
Example 10: In the figure shown, find the equivalent capacity.
Solution: Between a and b, there
is a parallel module that simply add up. C_{ab} = 60.0μF. Then, C_{ab} and C_{bc} are in series and their reciprocals add up to give the reciprocal of C_{ac}. 1/C_{ac} = 1/C_{ab} + 1/C_{bc} = 1/60 + 1/20 = 1/15. C_{ac} = 15.0μF. 
Fig. 13 
Example 11: In the figure shown, find the voltage across and the charge accumulated in each capacitor.
Solution: From the top figure: C_{ab} = 12.0 + 15.0 = 27.0μF ; and from the bottom figure: 1/C_{ac} = 1/C_{ab} + 1/C_{bc } ; 1/C_{ac} = 1/27 +1/13.5 ; C_{ac} = 9.00μF. This is the overall capacity that the 18.0V battery faces. Since Q = CV; thus , Q = (9.00μF)(18.0V) = 162 μCoul. This means that each capacitor in the bottom figure accumulates 162μC of charge (They are in series). Knowing their capacities, we can calculate their voltages. Q_{ab} = C_{ab}V_{ab} ; V_{ab} = 6.00 Volts. (across C_{1} & C_{2}) Q_{bc} = C_{bc}V_{bc} ; V_{bc} = 12.0 Volts. (across C_{3}) Going to the abportion of the top figure, we may find how the two parallel capacitors divide the 162μC of charge. They divide it as (12/27) and (15/27) proportions. Go to the next column.

Fig. 14 Q1 = (12/27)(162μC ) = 72.0μC. Q2 = (15/27)(162μC ) = 90.0μC. Of course, Q3 = 162μC.

Energy Stored in a Capacitor:
Note that the product QV has unit of energy. Q is in Coulombs and V is in Joules/Coul.. The product QV has unit of Joule in (SI). Keeping this in mind, let's calculate the energy stored in a capacitor that has charge Q on it causing a voltage V across it.
There are two other versions of this formula. Since Q = CV, we get U_{e} = (1/2)CV^{2}. Also, U_{e} = (1/2)QV. Verify.
Example 12: A 15μFcapacitor is connected to a 9.6V battery. Calculate (a) the charge accumulation and (b) the energy stored in it.
Solution: (a) Q = CV ; Q = (15μF)(9.6 V) = 144μC.
(b) U_{e} = (1/2)QV ; U_{e} = (1/2)(144μC )(9.6V) = 690 μJ.
Test Yourself 2:
1) A capacitor is a device that stores (a) kinetic energy (b) electric energy (c) elastic potential energy. click here.
2) The capacity, C, of a parallelplates capacitor is proportional to (a) the area of one of its plates, A (b) the the reciprocal of the gap between its plates, (1/d) (c) to the dielectric constant, κ of the the material between the plates (d) a, b, and c.
3) The dielectric constant, κ of the insulating material between the plates of a capacitor is (a) the ratio of the permittivity of that material, ε to the permittivity of vacuum, ε_{o} (b) such that we may write: ε =κε_{o} (c) both a & b. click here.
4) The value of ε_{o}, _{ }the permittivity of vacuum for the passage of the electric field effect, is equal to (a) 8.85x10^{12} Farad/meter (b) 8.85x10^{12} Coul.^{2}/(Nm^{2}) (c) 1/(4πk) where k is the Coulomb's constant. (d) a, b. &c. click here.
5) Capacity is also defined as (a) the chargetovoltage ratio of a capacitor (b) charge todistance ratio of a capacitor (c) chargetoenergy ratio. click here.
6) When an empty capacitor is connected to a battery, the very first voltage across the capacitor is (a) zero (b) exactly equal to the battery voltage (c) half of the battery voltage.
7) When an empty (deflated) basketball is connected to an air pump, the very first gauge pressure in the basketball is (a) zero (b) equal to the pump's or the compressor's pressure (c) half of the compressor's pressure.
8) When the capacitor in Question 6, is halfcharged, the voltage across it is (a) equal to the battery's voltage (b) equal to 1/2 of the battery's voltage (c) fluctuates. click here.
9) When the basketball in Question 7, is halfway filled, the air pressure in it is (a) equal to the pump's pressure (b) equal to 1/2 of the pump's pressure (c) fluctuates.
10) When the capacitor is fully charged after sufficient time has elapsed, the voltage across it (a) is almost equal to the battery's voltage (b) is zero because it does not accept any more charges (c) is neither a nor b. click here.
11) When the basketball in Question 7, is fully inflated to where the pump cannot inflate it anymore, the pressure in it (a) is equal to the pressure that the pump can generate (b) is zero because it does not accept any more air (c) neither a nor b.
12) The above questions lead to (a) the proportionality of charge, q, on each plate of a capacitor to the voltage, V, developed across it (b) the fact that capacity, C, is the proportionality constant (c) Q = CV. (d) a, b, & c.
13) The voltage across a 12μF capacitor is 5.0V. Each of its plates carry a charge of (a) 60.μCoul. (b) 2.4μCoul. (c) 0.
14) The charge on, and the voltage across a capacitor are 85μCoul. and 5.0 volts, respectively. Its capacity is (a) 425μF (b) 425 Farad (c) 17μF. click here.
Problem: Draw a battery and three parallelplate capacitors connected to it in series as shown in Fig.9. During the very first moments, suppose 5 trillion negative charges travel from the negative pole of the battery and distribute evenly over the nearest plate they can reach.
15) What happens to the other plate of that capacitor? (a) It receives 5 trillion electrons (b) It loses 5 trillion electrons (c) It becomes 5 trillion protons positive (d) b & c. click here.
16) Where do the repelled electrons of the first capacitor go? They (a) jump into air (b) go to the nearest plate of the middle capacitor and make it 5 trillion electrons negative (c) they return to the negative plate the same way they came in.
17) What happens to the opposite plate of the middle capacitor? It becomes (a) 5 trillion protons positive (b) 5 trillion electrons negative (c) Neither a no b. click here.
18) Is it correct to say that the third capacitor experiences the same process as the middle one? (a) Yes (b) No
19) What happens to the repelled electrons from the third capacitor? They (a) go to the positive pole of the battery and get absorbed by it (b) complete the flow of electrons in the circuit (c) both a & b. click here.
20) If we name the capacitor's charges Q_{1}, Q_{2}, and Q_{3}, then (a) Q_{1}= Q_{2}= Q_{3} (b) Q_{1}= Q_{2}+ Q_{3} (c) neither a nor b.
21) We may say the capacitors in series accumulate the same amount of charge. (a) True (b) False click here.
22) For capacitors C_{1} and C_{2} in series (C_{1 }> C_{2}) with a battery_{ }(a) Q_{1}>Q_{2} (b) Q_{1}< Q_{2} (c) Q_{1}= Q_{2}.
23) For capacitors C_{1} and C_{2} (C_{1 }> C_{2}) in parallel with a battery_{ }(a) Q_{1}>Q_{2} (b) Q_{1}< Q_{2} (c) Q_{1}= Q_{2}.
24) The equivalent capacity, C_{eq} for C_{1}= 25.0μF and C_{2} = 5.00μF connected in parallel is (a) 4.25μF (b)30.0μF (c) 125μF.
25) The equivalent capacity, C_{eq} for C_{1}= 25.0μF and C_{2} = 5.00μF connected in series is (a) 4.17μF (b)30.0μF (c) 125μF.
26) The equivalent capacity, C_{eq} for C_{1}= 15μF and C_{2} = 52μF connected in series is (a) more than 67μFy (b) less than 15μFy (c) equal to 67μF. click here.
27) Two capacitors C_{1}= 8.0μF and C_{2} = 16μF are connected in parallel to a 4.0V battery. The accumulated charges are: (a) 32μCoul. and 64μCoul. (b) 2.0μCoul. and 4.0μCoul. (c) neither a nor b.
28) Two capacitors C_{1}= 8.0μF and C_{2} = 24μF are connected in series to a 4.0V battery. The accumulated charges are: (a) 32μCoul. and 96μCoul. (b) 2.0μCoul. and 6.0μCoul. (c) 24μCoul. and 24μCoul.. click here.
29) The voltages across C_{1} and C_{2} above are (a) 3.0V and 1.0V (b) 4.0V and 4.0V (c) 24.0V and 24.0V .
30) The product QV has unit of (a) force (b) power (c) energy. click here.
31) When a capacitor is charged, it can give back the accumulated charge, Q, on it (a) as it keeps the same voltage V (b) as the voltage across it decreases with gradual charge loss (c) as the voltage across it increases with gradual charge loss.
32) Based on the previous questions, the energy stored in a capacitor, U_{e}, is the product (a) QV (b) (1/2)QV (c) 2QV.
33) Because Q = CV, the stored energy, U_{e }= (1/2)QV may be written as (a) U_{e }= (1/2)CV^{2} (b) U_{e }= (1/2)Q^{2}/C (c) both a & b.
34) The energy stored in a 60.0μFcapacitor when the voltage across it is 5.00V is (a) 1500μJ (b) 3000μJ (c) 750μJ.
35) The charge accumulation on the capacitor of the previous problem is (a) 300μCoul. (b)12.0μCoul. (c) 120μCoul..
36) If you now use U_{e }= (1/2)QV to calculate the energy again, you get (a) 1500μJ (b) 3000μJ (c) 750μJ. click here.