Two point charges of +1.0 μc and -2.0 μc are located 0.50 m apart. what is the minimum amount of work needed to move the charges apart to double the distance between them? (k = 1/4π ε0 = 8.99 × 109 n · m2/c2)

The minimum amount of work needed to move the charges apart to double the distance between them is 0.018 J.

We know to change the position of a charge from one point to another work done can be written as,

Also, we know that the potential energy can be written as,

Given to us

q₁ = +1.0 μC  = 1 x 10⁻⁶ C

q₂ =  -2.0 μC = 2 x 10⁻⁶ C

m

= 0.5 m

Substitute the values,

Hence, the minimum amount of work needed to move the charges apart to double the distance between them is 0.018 J.

Learn more about Work:

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W = 0.018 J

Explanation:

As we know that work done to change the position of charge is given as change in the potential energy of the system

here we can say

so here we have

so we can say

now plug in all values in it

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.