Electrostatics doubt- What is the force between........?
603730
The following refers to Coulomb force between two charges in a single uniform medium, not the two-medium situation described in  the question.\nAnswer 1 is helpful but the lack of exponent signs makes it misleading. The force is inversely proportional to the square of the distance, so the simplified formula is \nF = k * q1 * q2 / r^2 \nk is a constant = 8.99*10^9 N m^2 / C^2 (C is charge in Coulombs). Thus if q1 = q2 = 1 Coulomb, and q1 is separated from q2 by r = 1 meter, the force is 8.99*10^9 N or 2.01 billion pounds! \nThe above simplified formula is a contraction of the full formula (see ref.),\nF = q1 * q2 / (4 * pi * epsilon * r^2)\nin which k is replaced by 1/(4 * pi * epsilon), where epsilon is the product of epsilon0, the electrical permittivity of the vacuum (= 8.85 * 10^-12 c^2/(N m^2)), and kappa, the dielectric constant which = 1 for vacuum and is assumed in the simplified formula.\nFor two segments of distance as you specify, first solve for F1 and F2, each as a function of a particular distance (r or d - r) and dielectric constant. Then my inclination would be to try the sum-of-inverse square root solution which works for a similar situation, gravitational force F at distance R, computed from F1 and F2 at partial distances R1 and R2, where R = R1 + R2, \n1/SQRT(F) = 1/SQRT(F1) + 1/SQRT(F2)\nSorry I can't be more definite than that.
770512
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
769351
this question was on this yr's midyear exam, ill give u a clue.\nyou need to seperate the forces acting on the wire into two types \n1) acting horizontally (the force to the left equals the force to the right)\n2)acting vertically (the force to the bottom {or the normal force = 20 *9.8} equals the force to the top)\n\nequate them both and you should get the answer
935603
The net force is F2 - F1 = 25 - 19 = 6N.\n\nFrom F = ma, a = F/m = 6/5 = 1.2 m/s^2, toward the west because F2 > F1.  Also, we're ignoring friction.
343981
In two charged plates, other than between the plates (and the edge effects), are there any field line? \nYES! other field lines are between the outer sides of the plates(depending on the charges of the plates and the convention used for the direction of the flow). To prove this, just connect the outer sides of the plates by a wire through an amper meter and you will see the meter moving.\n\nWhy is the electric field constant everywhere in the space between parallel plates? \nThey are not unless the area of both plates is inifinte and that the two plates are parallel.  In this case, the 1) direction of the field is the same every where since by definition the plates are parallel and 2) the absolute value is constant.  If not, there will be a surface current and the current would make the entire electric field (its magnitude) to be constant over the entire surface.
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
1125356
Coulombs law\nThe force is inverse proportional to the square of their distance d.\nIf d 3x bigger, then F 3^2 times smaller\nThe force is proportional to each charge.
791491
The maximum possible net force is 125N, if the two forces are pushing in the exact same direction.\n\nForce is a vector quantity, which means that any measurement of force includes both a magnitude (size, the number) and a direction (the way that the force is pushing).  The simplest way to indicate direction is with a positive or negative sign, which only gives us two options: left/right, forward/backward, up/down, etc.\n\nIn your question, no direction is specified for either of the forces, but we can figure out the minimum and maximum possible net force rather easily.  The minimum net force would result when the two forces are pushing directly against one another:  100 -25 = 75N (toward the direction that the 100N was pushing).  Similarly, the maximum net force would result when the two forces are pushing together in the same direction:  100 +25 = 125N.\n\nIn general, then, the maximum possible net force (in cases where you are not given directions with the forces) can be found by adding the magnitudes of the forces together, as if they were all pushing in the same direction.
986389
Energy = force * distance\n= 100 N * 6 m = 600 J\n\nAnswer is d
667015
is that 1 cm or 10 cm?\n\nF1,2 = 9 x 10^9 x charge1 x charghe2 / distnace^2\n\nif distance is 1 cm = 10^-2 m\n8.5 micro C = 8,5 x 10^-6 C\n\nF1,2= 9 x 10^9 x 8,5 x 10^-6 x 8,5 x 10^-6 / 10^-4 = 6500 N\n\nif distance is 10 cm = 10^-1 m\n\nF1,2 =  9 x 10^9 x 8,5 x 10^-6 x 8,5 x 10^-6 / 10^-2 = 65 N\n\nthey will repulse each other with this force.\n\nbut also the other charghe is there, so you have to add vectorially the two forces to find out the resultant force on one charge.\n\nF^2 = (F1,2)^2 + (F1,3)^2 - 2 x F1,2 x F1,3 x cos 120\nF^2 = 3 x 65^2\nF = 65 x 1.73 = 112.58 N\n\nor F^2 = 3 x 6500^2\n   F = 6500 x 1.73 = 11258.33 N
210784
It's Gauss, not gows.\n\nE is the electric field on the surface (from whatever source, but you can use just from charges inside and you will get the same result).
269565
The electric potential is the potential energy per unit charge, so \n\nV = PE/q\n\nAlthough strictly speaking, the above should be written as the change in electric potential = change in potential energy/charge.  That's because there is no absolute electric potential, or potential energy.  It depends on what your reference is.\n\nNow, from that definition a number of equations arise.  It depends on the context of your problem.  If you have an electric field, and you are moving a charge through it, your change in potential is given by\n\nV = E d \n\nThe above equation follows from the definition of electric field as the electric force/unit charge, and the equations for Work = Force x distance.\n\nIf you have a circuit that obeys Ohm's law, then\n\nV = I R where I is the current in the circuit and R is the resistance.
813028
Well, as you've written it, it's a matter of differentiation the first equation, which gives you the second, as long as you take the limits of the integral properly.\n\nObjects in a potential field have an energy due to that field, dependent on their position, x. They will tend to move in a direction that lowers their energy, and experience a force due to the field that pushes them in the right direction. Hence, after some maths, the second equation - the force experienced due to a potential field is equal to the rate of change of that potential with respect to position - if the potential field changes rapidly, the force will be stronger. The minus sign means that the force moves you from high energy to low energy, ie, U decreases as the force moves you.\n\nThe first equation states much the same thing, the change in potential energy is from the force acting over a distance. This is usually how potential energy is defined, in case you're wondering where is came from.\n\nWrite Delta u as U(0) - U(x): the change in energy, moving from your starting point x=0 to some other point. Then the integral is\n\nS F(x) dx, with upper limit of x and lower limit of 0.\n\nDifferentiate both sides. The "U" side gives you -dU/dx, because U(0) is a constant, and the "F" side gives you F(x).\n\nIt's hard to explain any further without diagrams and stuff. Most first year college physics texts have more detailed explanations, better than I can. Hope this helps.
336359
The force of attraction between moon and earth is 2.3 x 10^ 20 newton (very strong);\n\n between earth and a molecule of oxygen~ 5x 10^ (-25) newton (weaker);\n\n between two oxygen molecules touching each other ~2x10^ (-42) newton; (weakest).\n\nThe electrical attraction between a hydrogen atom’s nucleus and its single electron is 9x 10^ (-8) newton, whereas the gravitational attraction is only 4x10^ (-47) newton.\n\nThe force of gravitation is negligible only in the case of interaction between elementary particles. \n\nONLY WHEN IFTERACTION BETWEEN ELEMENTARY PARTICLES ARE CONSIERED GRAVITATIONAL FORCES ARE WEEK.\n\nHence it is not correct to call gravitational force as week force.\n\nThe force of attraction between moon and earth is 2.3 x 10^ 20 newton (very strong);
367853
F = m*a \n\nm = 2 kg (two masses; two strings)\na = gravitational constant which happens to be 9.81 m/s^2 on the surface of the earth (or 32.2 ft/s^2 in EE units)\n\nsum the forces in the y-direction:\n\nT = F = 2 * m * a = 2 N\n\n(assuming massless strings)
40451
well figure out the net force with F=ma and just subtract. like just say 93 x .26 is the net force, the force thats acually being used for motion. so you subtract this from 36 to see what force is opposing this. so the total equation is:\n36 - (93 x .26)