Two particles, 1 and 2, each with charges 2q and 3q, are placed on a smooth floor at a distance d. A third particle is to be clamped to the floor in such a way that particles 1 and 2 will be in equilibrium on the floor only under the electric force. Where should particle 3 be clamped to achieve this?
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00:00Hi friends, dream as high as possible and achieve those dreams by studying.
00:07The Coulomb force is a force that arises from the interaction between two electrically charged particles.
00:14What if there are several electrical charges in a room?
00:19Of course, each particle will experience several forces.
00:24In this case, we are asked to determine the location of the electrically charged particles so that the other electrically charged particles are in equilibrium.
00:39As a visual illustration, on a smooth surface there are two electrically charged particles.
00:46We can call these particles particle 1 and particle 2.
00:53The electric charge from the particles with the same sign causes the particles to move away from each other.
00:59Especially if the floor surface is slippery.
01:03This is not the condition written on the problem sheet.
01:10We will attach another particle at a point such that particle 1 and particle 2 are in force equilibrium or do not move at all.
01:22Particle 3 will be attached to the floor surface so particle 3 cannot move at all.
01:29Now, we are asked to determine where the point of particle 3 is located.
01:37In equilibrium conditions, the resultant force acting on particle 1 and particle 2 is zero.
01:43How about we start analyzing particle 1.
01:47Particle 1 will move away from particle 2 because their charge signs are the same.
01:52So the force F12 will go to the left.
01:57Particle 1 will approach particle 3 because their charge signs are different.
02:02The force F13 will go to the right.
02:08The resultant force on particle 1 is zero, F13 minus F12 is equal to zero.
02:14In general, the direction of the vector to the right is positive.
02:18But unfortunately we do not know the distance between the particles.
02:23Let's assume the distance between particle 1 and particle 3 is x and the distance between particle 1 and particle 2 is d.
02:34This is quite a useful suggestion.
02:36In calculations, the value of electric charge is always positive regardless of its sign.
02:43This is because the sign of the charge is represented by the direction of the force drawn.
02:51We can shift the negative sign factors to the right.
02:56We can cancel out the K and Q factors.
03:00This is equation 1.
03:05Now we will analyze the force equilibrium for particle 2.
03:09Particle 2 tends to move away from particle 1.
03:12The force F21 will be to the right.
03:17Particle 2 tends to move towards particle 3.
03:22The force F23 will be to the left.
03:27Particle 2 is also in force equilibrium, F21 minus F23 equals zero.
03:36I think we have minimized the formula for the Coulomb force.
03:41Move the factors with negative signs to the right side.
03:46Similar to the previous process, cross out the factors K and Q.
03:51This is equation 2.
03:55Using these two equations, we will try to find the value of x.
04:00Up to this point, we only need math skills.
04:05Write equation 1 as an equation for Q.
04:10Next, substitute the value of Q into equation 2.
04:16It turns out that on both sides there is a factor of Q squared over d squared.
04:25Just cross out that value.
04:29From here, 3x squared is equal to 2 times the square of d minus x.
04:37This factor is d squared minus 2dx plus x squared.
04:44We will get a quadratic equation.
04:49Maybe we still remember the ABC formula.
04:54The square root of 24d can be written as 2d root 6.
05:01There are two possible values of x.
05:07I think this value is impossible because particle 3 will be to the left of particle 1.
05:14If so, the direction of the force acting on particle 1 will be towards the left.
05:20The resultant force on particle 1 will not reach equilibrium because both forces acting on particle 1 are both towards the left.
05:30So the answer is square root 6 minus 2 times d.
05:38This value is 0.45.
05:43So particle 3 must be inserted 0.45 times the length d to the right of particle 1 so that particles 1 and 2 are in equilibrium.
