Chapter 19

Electric Potential:

The discussion of electric potential is important because we are always looking for convenient sources of energy.  Since any two electric charges exert a force of attraction or repulsion on each other, if one charge moves in the field of the other through a distance Δr under an average force F, the work done is equal to F∙Δr.  This means that if there is just a single charge alone in the entire space, there is potential.  When a second charge moves in the field of the first charge, then work and energy actualize and we think of potential energy.  The electric field of point charge q1 at a distance r is  E = kq1/r2.   If another charge like q2 moves toward or away from q1, the change in the potential energy of q2 within Δ is  ΔP.E.= FΔr = q2E∙Δr.  We may write:

ΔP.E. = q2( kq1/r2 )∙Δr = q2E∙Δr.

Dividing both sides by q2 gives us the potential energy change per unit charge.

ΔP.E. /q2 = ( kq1/r2 )∙Δr.

From the units point of view,  the product of Δr and 1/r2 simplify and the result (unit-wise) is equivalent to 1/r .  The left side, the energy per unit charge is called the electric potential shown by symbol V.  The right side is equivalent to kq1/r.  This can be derived by using integration and calculus.

From the above, we accept without proof that the electric potential  V of a point charge q1 at a typical point P in space at a distance r  from it is given by :

This formula shows that as we move away from q1, the potential approaches zero; in other words, as r → ∞,  V→ 0.

Now if another charge like q2  is placed at Point P a distance  r  from  q1 , then  q2  finds a potential energy equal to

Consequently, we can write                               P.E. Vq2

The reason for thinking of a quantity such as potential is that, if for example, q1and q2are both positive, there is a repulsive force between them and if q2 is free to move, it can do some work for us or release some potential energy as it moves farther fromq1.   If we examine the unit of V, we see that it has units of (energy per charge) or in SI (Joules / Coulomb).  Let's do this examination.  Also let [  ] denote "the unit of ".

(Joule / Coul )  is called ( Volt ).  1 Volt means (1 J/Coul. ).  The unit is verified.  That is why  V is called potential.

1 volt is the potential of a point around charge (+q) such that if 1C  is placed at that point, 1 Joule of work is produced as (+q) repels the 1C charge to infinity.

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 q1 = 25.0nC at distances of 1.00m, 2.00m, and infinitely far from it..

Solution: V1= kq1 / r1 =  (9.00x109)(25.0x10-9) / (1.00) = 225 J /Coul. or  225 Volts.

V2 = kq1 / r2 (9.00x109)(25.0x10-9) / (2.00) = 113 J /Coul. or  113 Volts.

V3 = kq1 / r3 (9.00x109)(25.0x10-9) / ( ) =

Example 2: Calculate the potential energy that another charge q= 5.00nC possesses when placed at each of the three different points of the previous example.

Solution:  P.E.1 = V1(q2) = 225(J/Coul.) * (5.00x10-9 Coul.) = 1130 nJ

This means that it takes 1130 nJ of energy to push a 5.00-nC of positive charge from infinity to a distance of 1.00m from charge q1 that is also positive.

P.E.2 = V2(q2) = 113(J/Coul.) * (5.00x10-9Coul.) = 565 nJ

It takes 565 nJ to push a 5.00-nC charge from infinity to a distance of 2.00m from q.

P.E.3 =  V3(q2)  =  (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 q1.

Example 3: How much energy is needed to place 1.00-μCof 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 is in space far away from other electric charges as well.

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 (W1 = 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 done is equal to the change in P.E. of the 2nd charge in the field of the 1st charge.  It is equal to:

 W1 = 0.  W2 = kq1q2 / r = [(9.00x109)(1.00x10-6)(1.00x10-6) / (0.250)  ]J  = 36.0 mJ  3. The 3rd charge faces resistance from both the 1st charge and the 2nd    charge.  W3  = {kq1 q3 / (0.250)} +{kq2 q3 / (0.250)}= 36.0 mJ +36.0 mJ  =  72.0 mJ  Finally,  Wtotal  =   W1 +W2 +W3   = {0 + 36.0  +72.0} mJ   = +108.0 mJ

1) The electric field strength E1 of a point charge q1 at a distance r is (a) E1 = kq1/r2.  (b) E1 = kq1/r.  (c)  E1 = kq1/r3.

2) The electric force F of field E1on charge q2 is (a) F = E1q2.     (b) F = {kq1/r2}q2.     (c)  both a & b.      click here

3) The electric potential V1 of a point charge q1 at a distance r is (a) V1 = kq1/r.  (b) V1 = kq1/r2.  (c)  V1 = kq1/r3.

4) The potential energy P.E. of point charge q at points in space where the potential is V1 because of electric field magnitude Eis  (a) P.E. =Vq.      (b) P.E. = (kq1/r)q2.     (c) P.E. =  kq1q/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 force-field formula F = Eq is true (a) only if E is caused by a point charge.  (b) only if E is uniform and caused by a parallel-plate capacitor.  (c) whether E is that of a point charge or the field between the plates of a parallel-plate capacitor.

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 potential, V at a point is (a) "V = K.E. /q2 at that point."     (b) "V =F/q2 at that point."    (c) "V = P.E./q2 at that point."      click here

9) The electric potential energy, P.E. is (a) P.E. =  Vq2.   (b) P.E. = Vq22.  (c)  P.E. =  (K.E.)q2.

10 ) The reason for using q2 in the above questions instead of just q is that (a) qis the charge that is placed in the field of q1.  (b) the potential, V,  field, E, and force, F, in the above formulas are caused by charge q1.  (c) both a & b.

11) If instead of q2 , we simply use 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 another charge .  (c) both a & b.

12) The Metric unit of E, the electric filed, is (a) Coul./m .   (b) N/Coul..  (c) N/m .       click here

13) The Metric unit of V, the electric potential, is (a) Coul/s.  (b) Joules/Coul.  (c)  Joules/m.

14) The potential , V, at 9.0m from a +25μCoul. charge is (a) 2778 J/Coul..  (b) 25000 J/Coul..    (c) 0.

15) If  +1.0Coul. of charge is placed at 9.0m from the charge in Question 14, it finds a potential energy of (a) 2778J.  (b) 0.0J.   (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) 0.0.     (c) - 25000J.

17) The reason why the answer in Question 16 is negative is that (a) the negative charge placed at 9.0m will be pulled by the 25.0μC charge and will release energy as it moves toward it.  (b) the charge actually loses energy.  (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, one corner has that 4.0μC charge, the energy given to another 4.0μC charge to place it at a 2nd corner is  (a) -0.144J.  (b) 0.072J.  (c)  0.072J.      click here

23)  In Question 22, to place another 4.0μC charge at the 3rd corner, it takes (a) 0.144J.    (b) -0.144J.    (c) 0.072J.

24)  Formulas  V1 = kq1/ and  P.E. =  kq1q/r  apply (a) to point charges only.  (b) surface charges only.  (c) both a & b.

25) In the space between a parallel-plate capacitor, electric field, E is constant.  Potential energy varies with distance from each plate.  The way P.E. varies with (x), the distance from one of the plates, is (a) proportional to x.  (b) proportional to 1/x.  (c)  proportional to x2.      click here

Parallel-Plates Capacitor:

Two parallel and metallic plates separated by an insulator form a capacitor that can store electric energy.  If two flat sheets of aluminum foil sandwich a thin sheet of paper, a 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 negative.  This negative side repels equal number of electrons from the other side and causes the other side to become positive.  The repelled electrons flow toward the positive pole of the battery where they get absorbed by it.  This process causes the battery to find new poles: the plates of the capacitor.  One difference is that the new poles have more charges accumulated on them.  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.  If accumulation exceeds a certain value, electric discharge takes place via a spark through the insulator.  The capacity of the plates is obviously dependent on the area of each plate.  Of course the assumption is that both plates are made equal and completely face each other.  The thinner the insulating material ( called the dielectric) the stronger the electric field between the plates and therefore the greater the capacity.  The Capacity of a parallel-plates capacitor is given by

In the above formula, is the area of each plate and d is the gap between the plates or the dielectric thickness. The quantity ( ε) is called the Permittivity of the material for electric field transmission through it.  The permittivity of vacuum (free space)is shown by (εo) .  These two quantities are related by  ( ε = κ εo) where  κ 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 gap between the plates), the capacity increases by a factor of 5.4 compared to vacuum or almost air.

Example 4: Calculate the capacity of a parallel-plate 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) = 17.5x10-9F = 17.5 nFNote: 1 Farad of capacity is a very large capacity

Example 5: Calculate the area of each plate of a 1.00-Faradparallel-plate capacitor with an air gap of 0.0500mm.

Solution:  Solving the capacity formula for (A), yields:   A = Cd .  Substituting yields:

A = (1.00 F)(0.0500x10-3m) / (8.85x10-12 F/m) = 5.65x106m2

This means a square of side 2380m = 1.48 miles  ( A capacitor 1.48mi by 1.48mi ?)

Charge-Voltage Formula:

For a capacitor, the charge on the capacitor, Q, and the voltage, V, it generates are related by the formula:

The capacity of a capacitor is 1F  if it can hold a maximum charge of  1Coul.when connected to a voltage of 1V.  Majority of capacitors are of very small capacities because with the normal voltages of a few 10 or 100 volts they can only hold charges of micro- or nano-Coulombs magnitudes.  Farad =  Coul. / Volt

Example 6:  Calculate the capacity of a capacitor that holds at most 30.0μof charge when connected to a 12.0V battery.

Solution:  C = (Q/V) = (30.0 / 12.0)    (micro-Coul./ Volts)        or        C = 2.50 μF.

Example 7:  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 charge-voltage ratio is C = Q/V.  The more the accumulated charge, the greater the voltage it makes.  Capacity C is constant, anyway.   C = (60.0μC) / ( 7.50 volts)  =   8.00μF  (micro-Farads).

Connection of Capacitors:

It is sometimes necessary to have a capacitor of a certain capacity that is not available in the lab.  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 of connection:

Fig. 1                                                        Fig. 2

 Series:The battery voltage must equal the sum of voltages across the three capacitors.  We may write:  Vtotal  = Vab + Vbc + Vcd    (1)  If 2 electrons flow to the left of C1, they repel 2 electrons from the right plate of C1 making its right plate positive.  Those repelled electrons move to the left side of C2 making it negative while repelling 2 electrons from the right side of it making its right side positive.  The same happens to C3.  The repelled 2 electrons from the right of C3 will be absorbed by the positive pole of the battery and the current is complete.  Of course saying "2 electrons" is just an example.  In reality some 1013 or 1014 more or less electrons might flow.  We end up with equal charges on capacitors that are in series. 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 = CeqV from which V = Q / Ceq. For each capacitor we may write its own charge-voltage relation. Vab = Q / C1, Vbc = Q / C2  ,  Vcd = Q / C3.  Using (1), results in   Q / Ceq  =  Q / C1  +  Q / C2   +  Q / C3    Dividing through by Q, yields:     1 / Ceq  =  1 / C1  +  1 / C2   +  1 / C3 Parallel:The total charge Qtotal  that leaves the battery distributes over the three capacitors such that  Qtotal  = Q1 + Q2 + Q3.  If capacities C1, C2, and C3 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 C1, 6 will flow to C2, and 8 will flow to C3 and settle on their left plates.  Equal number of electrons will be repelled from the right side plates making them positive.  The repelled 18 electrons will be absorbed by the positive pole of the battery and the current is complete.  This is just an example, in reality some 1013 or 1014 more or less electrons might flow.  Qtotal  = Q1 + Q2 + Q3. Equivalent Capacity: Using Q = CeqV for the equivalent capacitor as well as the individual capacitors, yields:    CeqV  = C1V  +  C2V  +  C3V Dividing through by V, yeilds:        Ceq  = C1  +  C2  +  C3

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 / Ceq  = 1 / C1 + 1 / C2 ; 1 / Ceq  = 1 / 30.0 + 1 / 6.00 ;  Ceq  = 5.00μF ;  Make sure you use horizontal fraction bars when verifying the solution. Example 9: A 30.0μF capacitor is in parallel with a 6.00μC capacitor.  Find the equivalent capacity.Solution: Ceq = C1 + C2  ;  Ceq  = 30.0 + 6.00 ;  Ceq  = 36.00μF

Example 10:  In the figure shown, find the equivalent capacity.

 Solution:  Between a and b, there is a parallel set that simply add up.  Cab = 60.0μF Then, Cab and Cbc are in series and their reciprocals add up to give the reciprocal of Cac.   1 / Cac = 1/Cab + 1/Cbc = 1/60 + 1/20 = 1/15       Cac =  15.0μF

Example 11:  In the figure shown, find the voltage across and the charge accumulated in each capacitor.

 Solution:  From the top figure: Cab = 12.0 + 15.0 = 27.0μF  ;  and from the bottom figure:  1/Cac = 1/Cab + 1/Cbc  ;  1/Cac = 1/27 +1/13.5 ;  Cac = 9.00μF.  This is the overall capacity that the 18.0-V 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. Qab = CabVab; Vab = 6.00 Volts (Same for C1 & C2) Qbc = CbcVbc ;   Vbc =  12.0 Volts  (For C3) Going to the ab-portion 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. 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 Coul., 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.

As the charge accumulation on a capacitor varies from 0 to Q, the voltage across it increases from 0 to V The electric energy Ueis not equal to QV.  In the charging process, a battery of voltage V pushes electric charge toward the capacitor.  Since charge varies from 0 to Q, it is like the work done by the battery is V times the average charge (0 + Q)/2 or simply Q/2.  Therefore, the energy stored is (1/2)QV.  We may write:

Ue = (1/2)QV.

There are other versions of this formula.  Since Q = CV, we get Ue = (1/2)CV2.   Also,   Ue = (1/2)Q2/C.

Example 12:  A 15μF-capacitor is connected to a 9.6-V battery.  Calculate (a) the charge accumulation and (b) the energy stored in it.

Solution:  (a)  Q = CV    ;    Q = (15μF)(9.6 V)  =  144μC

(b)  Ue = (1/2)QV    ;    Ue = (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 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, εo.  (b) such that we may write: ε = κεo.  (c) both a & b.      click here

4) The value of εo the permittivity of vacuum for the passage of the electric field effect, is (a) 8.85x10-12 Farad/meter.  (b) 8.85x10-12 Coul.2/(Nm2).  (c) 1/(4πk) where k is the Coulomb's constant.   (d) a, b. &c.      click here

5) Capacity is also defined as (a) the charge-to-voltage ratio of the capacitor.  (b) charge -to-distance ratio of a capacitor.  (c)  charge-to-energy 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 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 half-charged, the voltage across it is (a) equal to the battery's voltage.   (b) equal to 1/2 of the battery's voltage.  (c) fluctuating.      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) fluctuating.

10) When the capacitor is fully charged after sufficient time has elapsed, the voltage across it (a) is 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 accumulation, Q on a capacitor's plate to the voltage developed, V across it.  (b) the fact that capacity C, is the proportionality constant.   (c) Q = CV .   (d) a, b, & c.

13) The voltage across a 12-μF capacitor is 5.0V.  Each of its plates has 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 parallel-plate capacitors connected to it in series as shown in Fig.1 above.  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? (a) They jump into air.  (b) They go to the nearest plate of the middle capacitor and make it 5 trillion electrons negative.  (c) the return to the negative plate the same way they came in.

17) What happens to the opposite plate of the middle capacitor? (a) It becomes 5 trillion protons positive.  (b) It receives 5 trillion electrons.  (c) Neither a no b.      click here

18) Is it correct to say that the third capacitor experiences the same process as the middle on? (a) Yes.  (b) No.

19) What happens to the repelled electrons from the third capacitor? (a) They go to the positive pole of the battery and get absorbed by it.  (b) They complete the flow of electrons in the circuit.  (c) both a & b.      click here

20) If we name the capacitor's charges Q1, Q2, and Q3,  then (a) Q1= Q2= Q3.  (b) Q1= Q2+ Q3.  (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 C1 and C2  (knowing C1>C2) in series with a battery,  (a) Q1>Q2.  (b) Q1< Q2.  (c) Q1= Q2 .

23) For capacitors C1 and C2  (knowing C1>C2) in parallel with a battery,  (a) Q1>Q2.  (b) Q1< Q2.  (c) Q1= Q2.

24)  The total capacitance, Ceq for C1= 25.F and C2 = 5.00μF connected in parallel is (a) 4.25μF.  (b)30.0μF. (c) 125μF.

25) The total capacitance, Ceq for C1= 25.F and C2 = 5.00μF connected in series is (a) 4.17μF.  (b)30.0μF. (c) 125μF.

26) The total capacitance, Ceq for C1= 15μF and C2 = 52μF connected in series is (a) more than 67μF.  (b) less than 15μF.  (c) equal to 67μF.      click here

27)  Two capacitors C1= 8.0μF and C2 = 16μF are connected in parallel to a 4.0-V battery. The accumulated charges are:(a) 32μCoul. and 64μCoul.   (b) 2.0μCoul. and 4.0μCoul..  (c) neither a nor b.

28)  Two capacitors C1= 8.0μF and C2 = 24μF are connected in series to a 4.0-V 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 C1 and C2 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 fully 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, Ue, is the product (a) QV.  (b) (1/2)QV.  (c) 2QV .

33) Because Q = CV, the stored energy,  U= 1/2QV may be written as (a) U= 1/2CV2. (b) U= 1/2Q2/C.  (c) both a & b.

34) The energy stored in a 60.0-μF-capacitor 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) 300.μCoul. .  (b)12.0μCoul. .  (c) 120μCoul. .

36) If you now use U= 1/2QV to calculate the energy again, you get  (a) 1500μJ.  (b) 3000μJ.  (c) 750μJ.       click here

Problems:

1) Charges q1 = 60.0pC and  q2 = 40.0pC are placed at the Origin and Point P( 20.0cm, 0.0), respectively.  Find the potential due to these two charges at the following points:    (a)  Point A( 0.0, 15.0cm),      (b) Point B(-30.0cm, 0.0),     and  Point C(0.0, -45.0cm).

2)  (a) How may electrons must be accumulated at a point so that the potential they create at a distance of 1.0mm from that point is 1.0Volt?  What is the voltage( or potential) at (b) 1.0cm from such electrons, and (c) at 1 micron? Note that 1micron = 1μm = 10-6m.

3) Calculate the voltage (electric potential) that the proton in hydrogen atom generates at 0.530Angstrom from it.  This is the smallest possible radius for the electron in hydrogen atom to orbit about the proton. 1 angstrom shown as 1Å = 10-10m or (one ten-billionth of a meter).

4) Calculate (a) the energy of a 10.0μC charge in the field of another 10.0μC charge placed at 90.0cm ( 1 yard) from it, and (b) the energy it takes to lift a 101.9-gram mass to a height of 1.00m above the floor with g=9.81m/s2.

5) In a circuit, 25.0mC of positive charge is flowing from the positive pole of a 40.0-V battery toward its negative pole.  This means that a 25.0mC of charge is placed in a uniform electric field of potential 40.0Volts.  Find the potential energy of the charge while moving from the positive pole toward the negative pole.

6) When you start your car in a hot summer and the engine is warm (low viscosity oil), the current pulled from the battery could be on the low side and about 40Amps (40Coulombs per second).  (a) How much energy is carried by 40C of charge in the uniform electric field of a 10-V battery?  Of course, this amount of energy is per second. When very cold (high viscosity oil), the current can be as high as 100Amps (100Coulombs per second).  (b) How much energy is carried by 100C of moving charge in the uniform electric field of an 9-V battery?

7) In the uniform electric filed between two parallel metallic plates that are connected to a 27-V battery, calculate (a) the electric potential, (b) the potential energy of a 1.0C charge,  (c) the P.E. of a 1.0μC,  (d) the P.E. of a 1.0pC, and (e) the P.E. of an electron.

8) (a) How much potential V does the proton in hydrogen atom generate at points that are at 0.53Å from it?  Note that nucleus is a point charge and its electric field E as well as potential V are not uniform and vary with distance. (b) How much P.E. does an electron attain when moves at such close distance to the nucleus (proton)?  (c) How many of such electrons are needed for their energy to add up to -1.0J?

9) How much energy is needed to place -1.00nC of electric charge at each corner of a square 0.450m long on each side?  Suppose the square is far from other electric charges.

10) Calculate (a) the capacity of a capacitor that holds at most 50.0nC of charge when connected to a 12.5V battery.  (b) What maximum charge does it hold when connected to a 25.0-V battery?

11) When a parallel-plate capacitor is 1/3 charged, it has 28.5μC of opposite charges on its plates and the voltage across it is 5.70 volts.  Find (a) its capacity,  the voltage across it when (b) 1/2 charged and (c) fully charged.

12) Capacitors C1, C2, and C3 with respective capacities of 10.0, 12.0, and 60.0 microfarads are connected in parallel and the two wire ends coming out of their combination has been connected to a 15.0-V battery for a long time.  Draw a diagram for the problem, then find (a) their equivalent capacity, C,  (b) the charge accumulation on each, respectively, and (c) the energy stored in each..

13) Capacitors C1, C2, and C3 with respective capacities of 10.0, 12.0, and 60.0 microfarads are connected in series and the two wire ends coming out of their combination has been connected to a 15.0-V battery for a long time.  Draw a diagram for the problem.  Then find (a) their equivalent capacity, C,  (b) the charge accumulation on each, respectively, and (c) the voltage across each capacitor.

1) 5.04Volts  2.52Volts,  1.93Volts

2) 690,000electrons, 0.10Volt, 1,000Volts

3) 27.2 Volts    4) 1.00J, 1.00J    5) 1.00J

6) 400J, 900J    7) 27V,  27J,  27μJ, 27pJ, 4.3x10-18J

8) 27.2 Volts,  -4.3x10-18J , 2.3x1017 electrons    9)108.4nJ

10) 4.00nF, 100.nC   11) 5.00μF,  8.55V, 17.1V

12) 82.0μF, (150.μC, 180.μC, 900.μC), (1.13mJ, 1.35mJ, 6.75mJ)

13) 5.00μF, (75.0μC, 75.0μC, 75.0μC), (7.50V, 6.25V, 1.25Volts)

Extra Examples:

 Example I:  In the figure shown, find the potential at Point B due to point charges q1 and q2.  Assume three significant figures. Solution: The distance from q1 to B is 10.0m, and the distance from q2 to B is (52 +52)1/2 m = 7.07m.  Since potential is a scalar quantity, the sum of the potentials at B becomes: Vtotal = V1B + V2B = kq1/r1 + kq2/r2  = 8.99x109 Nm2/C2[-20μC/10m + 50μC/7.07m]= 45600J/C.

Example II:  How many protons must be placed at the center of a sphere 12.0cm in diameter such that the potential of any point on the sphere's surface is1.00Volt?

Solution: Any point on such sphere is only 6.00cm from charge q that must be placed at its center.

Using V = kq/r with V = 1.00volt and r = 0.0600m, we get q = 6.67x10-12 C.

The number of protons needed to come up with this size charge is  6.67x10-12 C/ 1.6x10-19C = 4.17x107protons.

Example III:  Calculate the electric field magnitude that the proton in each hydrogen atom generates at points that are at a distance of r=0.53x10-10m from it.  This distance is the radius of hydrogen atom in its ground state when it is not energized by any external means.

Solution: The proton charge is q = 1.6x10-19C and r = 0.53x10-10m.  Solving for the electric field magnitude, we get:

E = kq/r2  = 8.99x109 Nm2/C2(1.6x10-19C ) / (0.53x10-10m)2 = 5.12x1011 N/C.