A point charge q1 = +5.00 is held fixed in space. From a horizontal distance of 6.00 cm, a small

step 1 Welcome to question number 23 .46. Chapter number 23, electric potential. We have given that the

A point charge q1 = +5.00 is held fixed in space. From a...

Key Concepts

Electrostatic Potential Energy

Electrostatic potential energy is the energy due to the configuration of two point charges. It is given by the expression k·q1·q2/r for two charges separated by a distance r. This concept is important in problems where the potential energy changes as the charges move relative to one another, and it is a key part of applying energy conservation to determine the distance (and hence the force) when the sphere's speed changes.

Coulomb's Law

Coulomb's Law quantifies the electrostatic force between two point charges. It states that the magnitude of the force is proportional to the product of the charges and inversely proportional to the square of the distance between them, with the constant of proportionality being Coulomb's constant. This concept is crucial when calculating the force that one charge exerts on another in an electrostatic system.

Newton's Second Law

Newton's Second Law relates the net force acting on an object to its mass and the resulting acceleration by the equation F = m·a. This principle is applied once the force due to electrostatic interactions is determined, allowing for the calculation of the acceleration experienced by the charged sphere.

Energy Conservation

The principle of energy conservation states that the total energy in an isolated system remains constant. In this context, the initial kinetic energy of the sphere and its electrostatic potential energy relative to the fixed charge add up to its total mechanical energy. As the sphere moves, this energy conservation is used to relate changes in its speed to changes in position (and hence the distance between charges), which then affects the electrostatic force acting on it.

Transcript

00:02 Welcome to question number 23 .46.

00:08 Chapter number 23, electric potential.

00:12 We have given that there are two charges, qn and q2 here.

00:18 And q1 has 5 microculum charge, so which is equal to 5 to 10 to minus 6 coulum.

00:24 And q2 has equals to 2 microculum charge.

00:28 So this is equal to 2 into 10 to minus 6 cullum.

00:31 Both are positively charged.

00:33 That means the force acting on q2.

00:36 Q1 by q2 is opposite or the repulsive force.

00:42 So we see that this is a repulsive force since they have same sign of charges.

01:01 So force by q1 on q2 is towards outward and force by q2 is towards this one.

01:11 So the question is find out the acceleration of the sphere at the instant when its speed is 25 meter per second.

01:19 So we are given that initially, when the q2 is at 6 cm distance 6 cm distance its velocity of the q2 is equals to 20 40 meter per second so this is yeah 40 meter per second so this particle is moving towards q1 and q1 is fixed here q1 is fixed here so q2 is moving towards the q1 with velocity in 40 meter per second when is q1 and q2 are 6 cm distance apart.

01:57 So the force is acting between q1 and q2 and q1 exert on q2 force in this way but the particle is moving this way so definitely the particle will deserate particle will disillate and velocity will decrease until it is stopped so the particle has velocity equals to 40 meter per second but the force is acting this way so the particle will have its velocity decreasing decreasing and reaches to some point where the velocity becomes zero but we don't need to care about the particle having velocity equals to zero because we are asked to find out the acceleration of the particle or the deceleration when the particle velocity is equal to 25 m per second so this has 40 meter per second so in the midway or somewhere it has velocity 25 meter per second so we need to find out this distance between q1 and q2 so let's say this is equal to d now so we have to find out the d here or let's say this is d prime yeah so sorry this is d so this is d prime this is equal to d i have assigned this six centimeter equals of d so use different let's use different symbol here d prime all right so we have used this formula here which is equals to k equals to half mv square kinetic energy formula and here the potential of the formula and here is the force formula between q and then q2 and here is the newton's third law which is equals to force equals to mass into exploration and this is work energy theorem whatever the energy of the particle q2 has initially will remain same here when the particle is at some final situation so this has some kinetic energy initially and some potential energy here and this will convert into the energy here and the potential energy so here the particle has higher velocity so the kinetic energy is higher and it decreases to k find them but the ui is lesser than uf so this is equal to this one so we will use this equality to find out the values here so this is conservation of energy formula so let's do the calculations now so we have k initially equals to half m v i square which is equal to take this mass half mass is your 4 into 10 dash to a minus 3 kg into 40 into square so this is that this is equals to 3 .20 jule now finding out the kf the final energy which is half mv final square which is this value so we have half 4 into 10 x0 minus 3 kg 25 square and this is equals to 1 .25 june now finding out the potential energy which is ui is equals to 1 by 4 pi epsilon q1 q2 over r the initial distance between them is 6 centimeter or 0 .06 meter so we have 9 into 10 to power 9 into q1 q2 charges so we have q1 here and q2 here over distance which is 0 .06 meter so using the calculator here i have found out the value this is equal to 1 .498 jule similarly we can find out the uf here but we don't know the uf since we don't know the uf since we don't know the distance between in q1 and q2 finally so we will replace this value as a variable here so this is q1 q2 over deep so this is your d prime now.

07:18 So using this quantity here we put the value to the k -i, ui and kf and uf here and find out the value of the d prime.

07:27 So we have uf equals to k i plus ui minus k f which is equals to 3 .2 plus 1 .25 minus 1 .498 and this value is equal to this one so we have 1 by 4 pi absonaut q1 q2 over d prime so d prime is equal to 3 .448 the value of this is equals to 3 .448 jule so putting this value we have 4 pi absent on power 1 minus sorry sorry sorry yeah so this yeah this is one minus and this is also one minus so we have q1 q2 let me write it again so your d prime this value so this comes out to be 3 .448 is equal to 1 by 4 pi epsilon q1 q2 over d prime so this gives us d prime equals to 1 by 4 pi epsilon q1 q1 q2 over 3 .448 which is equal to same as follows...

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