April 19, 2014

Search: physics (coulombs law)

Number of results: 116,236

What is the question on a) Coulombs law b) the charge equalizes IF the SPHERES are the same size. 6 microC each c. Again, Coulombs law. Notice you dont have to do much math on c). the difference between a and c is the numerator, in a) you have 20, and in c you have 36. So the ...
Saturday, January 26, 2013 at 4:22pm by bobpursley

You can do this... Coulombs law Newton's law of gravity.
Monday, February 2, 2009 at 9:41pm by bobpursley

These are all Coulomb's law F = k Q1 Q2/r^2 Just plug in but watch units 2.4*10^-6 Coulombs at .055 meters in prob 1 for example. k = 9*10^9 e = 1.6*10^-19 Coulombs, negative for electron, positive for proton
Sunday, March 20, 2011 at 8:58pm by Damon

Using Coulomb's Law, calculate the force of repulsion between a positive charge of 110 Coulombs and another positive charge of 220 coulombs separated by a distance of 2 meters. Use scientific notation. (Hint: use the proportionality constant)
Sunday, June 17, 2012 at 9:54am by LL

Using Coulomb's Law, calculate the force of repulsion between a positive charge of 110 Coulombs and another positive charge of 220 coulombs separated by a distance of 2 meters. Use scientific notation. (Hint: use the proportionality constant)
Sunday, June 17, 2012 at 9:21pm by LL

1) The curent is I = 5 Amps (Coulombs per second) According to Ampere's law, the magnetic field at a distance of R from the conductor is B = Uo* I/(2*pi*R) Make sure R is in meters. Uo is the permittivity of free space. 2) Same formula for B, but I = 3*10^20*e/(2*10^-3 s) ...
Wednesday, April 22, 2009 at 6:37pm by drwls

Use Coulombs law.
Sunday, February 7, 2010 at 7:51am by bobpursley

Coulombs law applies.
Monday, April 19, 2010 at 8:30am by bobpursley

F=kQ1Q2/d^2 coulombs law....
Tuesday, April 10, 2012 at 8:00pm by bobpursley

Use coulombs law.
Saturday, October 27, 2012 at 7:06pm by bobpursley

coulombs law... F=kq1*q2/r^2
Tuesday, April 10, 2012 at 8:01pm by bobpursley

F=kq^2/d^2 coulombs law q=sqrt( F*d^2/k)
Monday, November 26, 2012 at 9:29pm by bobpursley

What is Coulombs law? I will be happy to critique your thinking.
Friday, January 22, 2010 at 3:36pm by bobpursley

Physics 221
Have you considered coulombs law? E=kq/r^2
Tuesday, December 31, 2013 at 10:05pm by bobpursley

Columbs Law
I need help with the following questions from my homework. thanks! How does one coulomb of charge compare with the charge of a single electron? How is coulombs law similar to Newton's law of gravitation? How is it different?
Monday, November 5, 2007 at 8:13pm by Whit

Use coulombs law on each of the three charges, and add them as VECTORS.
Wednesday, February 24, 2010 at 10:20am by bobpursley

Use Coulombs law force=kQtop*Qbottom/d^2 solve for d
Tuesday, March 15, 2011 at 3:29pm by bobpursley

Physics - Coulombs Law
Two identical objects have charges 6.0*10^-6 and -2.0*10^-6 respectively. When placed a distance d apart, their force of attraction is 2.0N If the objects touched together, then moved to a distance of separation of 2d, what will be the new force between them. Ans: I tried the ...
Sunday, October 17, 2010 at 8:52am by Anonymous

Find the distance to the point from the origin. d=sqrt (x^2+y^2) working it in meters Then coulombs law.
Thursday, August 13, 2009 at 11:06am by bobpursley

Hardly. It seems to me Coulombs law and the Magnetic strength equation are very similar.
Monday, April 16, 2012 at 4:49am by bobpursley

Yes, +2 e coulombs. That's often written just +2. e = 1.6019*10^-19 coulombs
Monday, February 25, 2008 at 11:07pm by drwls

3.5 Amp = 3.5 Coulombs per second. So, multiply that by 0.8 seconds. That will be the answer, in Coulombs.
Friday, May 11, 2012 at 11:32am by drwls

It takes 96,485 coulombs of electricity to plate (deposit) 1/2 mole copper metal. Use that information to calculate coulombs. Then amperes x seconds = coulombs.
Saturday, May 1, 2010 at 11:05am by DrBob222

Coulombs Law of repulsion: F=k q1*q2/distance^2 change units to meters, coulumbs.
Sunday, February 6, 2011 at 6:08pm by Anonymous

(a) Use Coulomb's Law. In this case, the Q's are both equal to 2e, where e = 1.6*10^-19 Coulombs b) According to Newton's second law, acceleration = (Force)/(mass) where the mass is that of an alpha particle, which is very nearly equal to the mass of 4 protons
Saturday, March 14, 2009 at 3:57am by drwls

opposite charges, they attract. Use coulombs law to find the force. c. excells= chargeonCork/chargeononeElectron
Friday, March 7, 2014 at 2:06am by bobpursley

One electron has a charge of 1.6E-19 coulombs 16E-19 coulombs/e x #e = 18E-6 coulombs. Solve for #e.
Monday, September 17, 2012 at 11:49pm by DrBob222

How far apart are they? Is the charge in Coulombs? The force they exert on each other will be independent of the external E field. Use Coulomb's Law.
Monday, November 5, 2012 at 5:15am by drwls

Coulombs Law: force=k q2 q1 / distance^2 solve for q2
Thursday, February 21, 2013 at 11:03am by bobpursley

(Sphere area @ r) * (E field @ r) = 4 pi k q That is Gauss' Law in this situation, and reduces to: 4 pi r^2 * E(r) = 4 pi k q E(r) = k q /r^2 which you might recognize as Coulomb's Law. Plug in k and q values to get the E field in Volts/meter (or Newtons/Coulomb). Make sure r ...
Sunday, April 11, 2010 at 12:53pm by drwls

Use Coulomb's law for the magnitude. The nuclear charge is +65 e, where e = 1.6*10^-19 Coulombs. The direction is radially outward from the nucleus, everywhere.
Sunday, February 8, 2009 at 3:53pm by drwls

You may not be able to post for a few days but I think you can read responses. Here is how you do this problem. You want 7.30g Ni. 96,485 coulombs will deposit 58.7g/2 or 29.35g Ni. so you will need 96,485 coulombs x (7.35/29.35) = approx 24,000 coulombs but that's an estimate...
Wednesday, April 2, 2014 at 8:32pm by DrBob222

use diff eq. with coulombs law E = kq/r^2 r^ Solve dy/dx = E_net,y/E_net,x
Monday, July 1, 2013 at 2:18pm by Anonymous

What is the force exerted on 40 micro coulombs by 20 micro coulombs? (charges are separated by 80cm)
Wednesday, May 2, 2012 at 6:16pm by paul

5.5 A = 5.5 Coulombs/second 2 minutes = 120 seconds Multiply those two numbers to get Coulombs. 5.5 x 120 = ___ C
Tuesday, July 10, 2012 at 11:22pm by drwls

3.94 x seconds = coulombs You want to know seconds (and convert to hours). 96,485 coulombs will plate out 207.2/2 = 103.6g Pb. So how many coulombs do you need? That's 96,485 x 5.40/193.6 = ? Substitute that ? into coulombs above, solve for seconds and convert to hours.
Wednesday, July 24, 2013 at 10:43pm by DrBob222

2 more physics questions I need help on, please?
current in amps is coulombs of charge per second I = Q/t so Q = I t 4*1 = 4 2*.5 = 1 8*2 = 16 coulombs Q = i t so t = Q/i t = 2.5 /5*10^-3 = (1/2) * 10^3 = 500 seconds
Saturday, April 23, 2011 at 7:57pm by Damon

A point charge of +10 micro-coulombs lies at x = -0.39 m, y = 0 m, a point charge of -5.7 micro-coulombs lies at x = +0.39 m, y = 0 m, and a point charge of +14.9 micro-coulombs lies at x = 0 m, y = -0.3 m. A point charge of +10 micro-coulombs is moves from x = 0 m, y = +0.11 ...
Friday, February 10, 2012 at 8:32am by Mandaleigh

The current in Amperes (which is Coulombs/s)is 5.0. Multiply that by 240 s for the charge transfer, in Coulombs. Divide the answer by the electron charge (1.6*10^-19 C) for the number of electrons transferred.
Thursday, October 14, 2010 at 3:02am by drwls

Nothing to be lost about. Coulombs Law: force=kQ1*Q2/distance^2 I can check your work. Work this in the SI system of units.
Tuesday, April 24, 2012 at 5:49pm by bobpursley

Coulombs law: Forceopposing=kQ1Q2/distance^2 notice the force opposing is negative, meaning they attract each other, so the forces are each towards the other.
Friday, December 7, 2012 at 10:01am by bobpursley

physical science
What is Coulombs Law?
Tuesday, October 27, 2009 at 9:27pm by bobpursley

Use coulombs law.
Friday, January 21, 2011 at 10:31pm by bobpursley

Your attempt to cut and paste your assignment failed. Use Coulombs law: F=kQ1*Q2/distance^2 you need to be careful with units, you did not specify dimentsions. acceleration= force/mass
Monday, August 29, 2011 at 6:09pm by bobpursley

A conducting spherical shell with a total charge of +12.6 micro-coulombs has a point charge of -3.6 micro-coulombs placed at its center. The negative point charge will push the electrons in the spherical shell away from it. This will leave the inner shell with positive charge ...
Friday, February 10, 2012 at 8:30am by Mandaleigh

a micro is 10^-6 or one millionth so we are talking about -30 *10^-6 coulombs the charge on the electron is -1.6 *10^-19 Coulombs so N electrons * (-1.6*10^-19 coulombs/1 electron) = -30 *10^-6 coulombs so N = 30 *10^-6 /1.6*10^-19 that is (30/1.6) * 10^13 which is (3/1.6) *10...
Monday, January 7, 2008 at 6:04pm by Damon

Coulombs law leads to this: E=kq/r^2
Tuesday, August 24, 2010 at 10:20am by bobpursley

Use coulombs law, add the forces as vectors. You have a S force from the y charge, and a W force from the x axis charge.
Sunday, February 21, 2010 at 11:19pm by bobpursley

What is the electric field at a point midway between a -7.34 micro coulombs and a +4.49 micro coulombs charge apart 31.5m apart?
Tuesday, July 9, 2013 at 9:17am by Amalia

Current flow through a lamp is 1.3 A. How many Coulombs of charge are passed through the lamp in 2 minutes? Formula for coulombs and current? Not sure..
Sunday, February 3, 2013 at 12:17am by Physics fail

what is the electric field at the center of the square? q1 and q2 = -10 micro coulombs q3 and q4= 5 micro coulombs each side is 0.10 m. How do i set this up?
Thursday, July 7, 2011 at 6:18pm by Sarah (C)

That many electrons have total charge Q = 3.5*10^20*1.6*10^-19 = 56 Coulombs Require that I*t = 56 C where I is the current in Amps (Coulombs per second). I = 0.130 C/s Solve for t.
Tuesday, February 22, 2011 at 11:09pm by drwls

Physics please help
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro coulombs and q2= -5 micro coulombs? Attempt at solution: no idea but need to get this right our homework is graded
Wednesday, February 15, 2012 at 9:30pm by bel

physics 2
A charge of -6.8 micro-Coulombs is placed at x = 0, y = 0. A charge of -18.5 micro-Coulombs is placed at x = 0, y = +37 cm. Another charge of -12.9 micro-Coulombs is placed at x = +17 cm, y = 0 cm. What is the angle the total electric force on the charge at x = +17 cm, y = 0 ...
Friday, January 18, 2013 at 2:18pm by Jessica

Apply Gauss' law at a distance r = 0.526 m from the center of the sphere. You should have provided units for the charge. Are they Coulombs or microcoulombs? You also did not provide units for the distances. Unless you understand the importance of units in physics and ...
Wednesday, February 10, 2010 at 2:38pm by drwls

add up all the flux then use Gauss law total flux out = charge inside/eo where eo = 8.854 *10^-12 coulombs^2/Nm^2
Thursday, September 3, 2009 at 7:41pm by Damon

Calculate the magnitude and direction of the Coulomb force on each of the three charges connected in series: 1st - 6 micro Coulombs (positive charge) 2nd - 1.5 micro Coulombs (positive charge) 3dr - 2 micro Coulombs (negative charge) Distance between 1st and 2nd - 3cm Distance...
Wednesday, May 16, 2012 at 11:19pm by andrew

Calculate the magnitude and direction of the Coulomb force on each of the three charges connected in series: 1st - 6 micro Coulombs (positive charge) 2nd - 1.5 micro Coulombs (positive charge) 3dr - 2 micro Coulombs (negative charge) Distance between 1st and 2nd - 3cm Distance...
Monday, May 21, 2012 at 7:35am by andrew

No, the answer is not 1 N. Look up and apply Coulomb's law. Force = k Q1 Q2/R^2 The k is the Coulombs constant. The rest of the terms in the equaation are 1 in SI units in your case. This results in a very large value for the force, in Newtons
Sunday, November 11, 2007 at 11:50am by drwls

amperes x seconds = coulombs. coulombs/96,485 = ??mol electrons.
Saturday, April 19, 2008 at 11:52pm by DrBob222

amperes x seconds = coulombs 96,485 coulombs = 1 mol electrons
Sunday, April 20, 2008 at 12:25am by DrBob222

amperes x seconds = coulombs. 95,486 coulombs = 1 mol electrons.
Tuesday, April 22, 2008 at 7:33pm by DrBob222

Find coulombs, then amp x seconds = coulombs. How many grams Cu do you have that's deposited.
Monday, June 18, 2012 at 2:15pm by DrBob222

Convert 1.0 L H2 to grams H2 and convert that to coulombs required remembering that 96,485 coulombs will release 1 gram equivalent weight of H2. Then amperes x time(seconds) = coulombs. Solve for time in seconds.
Wednesday, June 5, 2013 at 10:58am by DrBob222

a) Use Gauss' Law for the surface charge density in coulombs/m^2. The front and back surfaces will have equal and opposite charge densities, since the net charge is zero. b) Multiply the answer in a) by the plate area, 0.25 m^2.
Friday, September 18, 2009 at 9:15pm by drwls

Use F=ma to find the weight of an electron, use a=9.8 m/s^2 use coulombs law F=k(Q1Q2)/r^2 rearrange to r^2=k(Q1Q2)/F use k=8.99 x 10^9 N m^2/C^2 and Q1=Q2=1.60x10^-19 C
Friday, November 26, 2010 at 6:21am by Dr Russ

coulombs = amperes x seconds 96,485 coulombs will deposit 63.54/2 g Cu and 32/2 g oxygen at STP
Friday, February 3, 2012 at 4:07pm by DrBob222

gen chem
A x seconds = coulombs. It will take 96,485 coulombs to plate (26.98/3) g Al. Coulombs = 96,485 x (7.437 x 3/26.98) = ?? Substitute into the first equation and solve for A. Time is 540 min x 60 s/min.
Saturday, May 8, 2010 at 6:56pm by DrBob222

physics coulombs law
a negative charge 5.0 X 10^-4 C exerts a repulstio force of 7.0 N .on second charge away 0.05 m away. what is the sign and magnitude of the second charge
Wednesday, December 5, 2012 at 4:37am by adithya

physics (coulombs law)
2m N/ 6 C = E in Newtons/coulomb (-2/6) (2m N) ==================================== for a) i assumed the charge is 0 'cos there's no charge placed before it correct? NO! E is force experienced by test charge of one coulomb
Saturday, December 8, 2012 at 9:48pm by Damon

The average current is (20 Coulombs)/1.1*10^-4 s = 1.8*10^5 Amperes. Multiply the charge transfer by the voltage difference to get the energy transfer. 20 Coulombs * 1.2*10^8 = 2.4*10^9 Joules
Monday, October 11, 2010 at 6:05pm by drwls

A point charge of +4.6 micro-coulombs lies at x = 0 and a point charge of -7.3 micro-coulombs lies at x = +3.3 m. Where on the x axis, is the total electric potential equal to zero? Answer in meters.
Friday, February 10, 2012 at 8:31am by Mandaleigh

5.00x 10^-2 mass with charge +0.75 micro coulombs is hung by a thin insulated thread. Charge of -0.9 micro coulombs is held .15meters directly to the right so that the thread makes an angle with the vertical. What is the angle and the tension in the string?
Tuesday, March 29, 2011 at 10:14pm by T

Two small conducting spheres are identical except that sphere x has a charge of -10 micro coulombs and sphere y has a charge of + 6 micro coulombs . After the spheres are brought in contact and then separated,what is the charge on each sphere ,in micro coulombs ?
Thursday, August 18, 2011 at 5:21pm by Troy

96,485 coulombs are required to produce 1/2 mole Cu(63.546g/2); therefore, x coulombs are required to produce 23.8 g Cu. There are 6.022 x 10^23 electrons in 96,485 coulombs.
Sunday, May 2, 2010 at 8:38am by DrBob222

You have to assume the small spherical insulator is a point charge. The force horizontal is given by coulombs law, so looking at the diagram, TanTheta=horizontalforce/vertical force. Of course, the vertical force is mg. Tension in the wire? SinTheta=horizonalforce/tension.
Sunday, March 23, 2008 at 10:36am by bobpursley

An infinite sheet with a surface charge density of -2.1 micro-coulombs/m2 runs parallel to the x axis. Another infinite sheet a surface charge density of +2.1 micro-coulombs/m2 runs parallel to the y axis. A charge of -4.1 micro-coulombs moves at a 45 degree angle towards the ...
Friday, February 10, 2012 at 8:33am by Mandaleigh

physics 2
The charge per unit length is -9*10^-9C/0.05m = =1.8*10^-7 C/m You will have to perform an integration of Coulombs law for the E-field along the x axis at x = 0.25. There will be no y component, due to symmetry. dE = k*(charge density)*x*dy/(x^2+y^2)^(3/2) Integrate from y=-0....
Sunday, September 19, 2010 at 7:38pm by drwls

C = q/V that is definition of capacitor, charge in coulombs over voltage in volts 5.2*10^-6 farads = 2.1 *10^-3 coulombs / v so V = 2.1 * 10^-3 / 5.2 *10^-6 = .404 * 10^3 = 404 volts
Tuesday, February 5, 2008 at 5:18pm by Damon

Use coulombs law to predict the magnitude of the force. Since the charges are opposite, there is an attractive force between them. Since the negative charge is on the bottom, the force on the negative charge is up.
Friday, April 13, 2012 at 1:58am by drwls

It takes 96,485 coulombs to deposit 1 equivalent weight of Al (27/3 = 9 grams). Therefore, it will take 96,485 x (25/9) to deposit 25.0 grams = ?? The amperes x sec = coulombs. You know sec (convert 5 min to seconds) and coulombs, calculate amperes.
Tuesday, April 20, 2010 at 5:07pm by DrBob222

a) Use Gauss' law. 4 pi R^2 E = N*e/(epsilono) N is the number of electrons and e is the electron charge in Coulombs. E is the electric field strangth in N/C "epsilon zero" = 8.8554*10^-12 C^2/N m^2 Make sure R is in meters b) Use the same equation, but use R + 0.1 m in place ...
Sunday, January 27, 2008 at 6:27pm by drwls

A charge of -6.5 micro-Coulombs is placed at the origin of a coordinate system. Another charge of -7.6 micro-Coulombs is placed at x = +0.29 m, y = +0.17 m. A third charge of +14 micro-Coulombs is placed at x = -0.29 m, y = 0 m. What is the magnitude of the total electric ...
Saturday, January 29, 2011 at 12:06pm by kelly

Calculate the coulomb force on q. They are all in a straight line q1, q2,q3. Q1:1.05 micro coulombs Q2:5.51 micro coulombs Q3:-2.15 micro coulombs The distance, r1, between q1 and q2 is 20.1cm. The distance, r2, between q2 and q3 is 15.3 cm. Also, calculate e coulomb force on q2
Tuesday, July 9, 2013 at 9:21am by Amalia

The final charge on each of the three separated spheres in part (c) is +3.0 µC. How many electrons would have to be added to one of the three spheres to make it electrically neutral? Divide 3*10^-6 Coulombs by 1.6*10^-19 Coulombs per electron. The will be the number of ...
Sunday, January 21, 2007 at 12:13pm by Ann

coulombs = amperes x seconds. C = 10A x 180 min x (60 s/min) = ?? It takes 96,500 coulombs to deposit 1 mol Cr/3 or 52.01/3 grams.
Friday, June 11, 2010 at 2:59am by DrBob222

Physics. Free electron density, please help!
Call the free electron density N, m^-3 e = electron charge , Coulombs V = electron drift speed I = current, Amps A = cross sectional area, pi r^2, m^2 I = N*A*e*V Solve for N in this case. Get the current I from ohms law. I = V/R
Wednesday, November 21, 2012 at 11:38pm by drwls

2H2O ==> 2H2 + O2 amp x seconds = coulombs. 96,485 coulombs will deposit 1 g H atoms or 2*96,485 coulombs will deposit 2g H2 molecules which will occupy 22.4L at STP. Coulombs for 1.0L will be 2*96,485 x (1.0/22.4) = about8615 (approximately) ,then amps x sec = about 8615, ...
Thursday, March 7, 2013 at 11:07pm by DrBob222

One coulomb = 6.24150965(16)10^18 electrons. 1.56*10^20 electrons = 1.56*10^20/6.24150965(16)10^18 coulombs =25 coulombs So the equation becomes: 25 coulombs = 2 coulombs * e^(0.05t) e^(0.05t) = 25/2=12.5 take ln on both sides: 0.05t = ln 12.5 = 2.526 t = 2.526/0.05 = 50.5 sec.
Tuesday, May 31, 2011 at 9:15pm by MathMate

Physics please help
I am not familiar with the terminology. How can a "potential" be an event or a time interval? Power = (current) x (voltage) The voltage is apparently 30*10^-3 V Convert the 3x10^-7 mol/(m^2*s) to coulombs/(m^2*s) 1 mole of Na+ ions contains 6.02*10^23 ions with a charge of 9....
Sunday, February 10, 2013 at 1:37am by drwls

physics (coulombs law)
a charge +6 C experiences a force of 2mN in +x direction. a)what was the electric field there before charge was placed ? b)describe the force which a -2c charge would experience if it is located at the point in place of the +6c for a) i assumed the charge is 0 'cos there's no ...
Saturday, December 8, 2012 at 9:48pm by lola

Have you read about Coulomb's law? If not, it's time you did. F = k*Q1*Q2/R^2 In your case, Q1 = -Q2 = 2.0*10^-6 Coulombs R = 5.0*10^-2 meters k is the "Coulomb constant", 8.99*10^9 N*m^2*/C^2 Solve for F in Newtons. The force will be attractive since the charges have opposite...
Wednesday, May 22, 2013 at 10:31am by drwls

amperes x seconds = coulombs. coulombs/96,485 = Faradays 1 Faraday will deposit 1 equivalent weight (1/2 mole) Zn. Subtract moles Zn from initial Zn concn and change to molarity.
Monday, February 23, 2009 at 10:58pm by DrBob222

physics 2
Three charges are placed on the y axis. One charge of -3.2 micro-Coulombs is placed at y = 0, another charge of +2.6 micro-Coulombs is placed at y = -0.29 meters, and another unknown charge is placed at y = +0.51 meters. The total force on the charge at y = 0 due to the other ...
Friday, January 18, 2013 at 2:19pm by Jessica

The voltage on A dropped from 10.70 to 6.40 V, or by 4.30 Volts. Capacitor A lost C(A)*deltaV = 1.264*10^-4 Coulombs of charge. That charge flowed to capacitor B, leaving it with 1.264*10^-4 Coulombs and a charge of 6.40 V C(B) = Q/V = 1.264*10^-4/6.4 = 19.8 uF
Wednesday, November 23, 2011 at 10:53pm by drwls

5.0 amps x 0.5 hr x 60 min/hr x 60 min/s = 9,000 coulombs. 2Br^- ==> Br2 + 2e You know that 96,485 coulombs will form 1 equivalent of Br2 which is 159.8/2 = 79.9 g Br2. You have 9,000 coulombs; therefore, how much Br2 will be formed? 79.9 x (9,000/96,485) = ?
Friday, October 26, 2012 at 1:16pm by DrBob222

amperes x seconds = coulombs coulombs/96,485 = Faraday's 1 Faraday will deposit 1 equivalent weight of Zn^+2 or 1/2 mole Zn^+2. I get 0.965 x 10 min x (60 sec/min) = ?? coulombs. That divided by 96,485 = 0.006 Faraday's and that will deposit 0.003 mole Zn^+2. How many moles Zn...
Thursday, February 19, 2009 at 6:58pm by DrBob222

what is the electric field at the position 20 cm from Q2 and 60 cm from Q1? Q2=40 micro coulombs Q1=20 micro coulombs Q2 and Q1 are separated by 80 cm
Wednesday, May 2, 2012 at 7:20pm by quinton

The plastic article must be one of the electrodes but it doesn't conduct electricity. Therefore, the graphite coasting serves that purpose. I would think a graphite rod or carbon rod would serve as the other electrode. amp x seconds = coulombs 0.5A x 300 s = 150 coulombs, 96,...
Sunday, March 10, 2013 at 6:39am by DrBob222

Use coulombs law for each force. Then, break up each of the forces into x,y components. You will have then six components. Now, look at each corner. Add the two x, and the two y, calculate the magnitude and direction from those two components. Repeat for each corner. Remember...
Wednesday, February 6, 2013 at 4:35pm by bobpursley

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