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Two positive charges 12×10^-6C and 8×10^-6 are 10cm apart. Find the workdone in bringing them 4cm together

Sagot :

Answer:

W = 5.76 J

Explanation:

Given that,

Charge 1, [tex]q_1=12\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=8\times 10^{-6}\ C[/tex]

The distance between charges is 10 cm.

Initial distance = 10 cm = 0.1 m

Final distance = 10 cm - 4 cm = 6 cm = 0.06 m

We need to find the work done in bringing the 4 cm together.

We know that,

Work done = change in kinetic energy

[tex]W=kq_1q_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})\\\\\text{Put all the values}\\\\W=9\times 10^9\times 12\times 10^{-6}\times 8\times 10^{-6}(\dfrac{1}{0.06}-\dfrac{1}{0.1})\\\\W=5.76\ J[/tex]

So, the required work done is 5.76 J.

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