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KIM [24]
2 years ago
6

A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

) to the point (x2, y2). Please use k for Coulomb's constant rather than writing it out as (1/4πϵ0).
How much work W is done by the electrostatic force on the moving point charge?
Physics
1 answer:
Scilla [17]2 years ago
3 0

Answer:

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Explanation:

Position of charge q₁ is (0,0)

Position of charge q₂ is (x₁,0)

So, the electric potential energy between the charges is given by :

U_1=k\dfrac{q_1q_2}{x_1}

Now the position of charge q₂ has been changes from (x₁,0) to (x₂,y₂). Now, electric potential energy between the charges is :

U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}

We know form the work energy theorem that, the change in potential energy is equal to the work done. Mathematically, it is given by :

W=-\Delta U

W=-(U_2-U_1)

W=(U_1-U_2)

W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Hence, the work done by the electrostatic force on the moving point charge is kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}}). Hence, this is the required solution.

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Answer:

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Explanation:

given,

mass of the box (m)= 6 Kg

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box is sliding downward so,

F - f = m a                        

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m g sinθ - μ N = ma                        

m g sinθ - μ m g cos θ = ma            

a = g sinθ - μ g cos θ                    

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

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