Two positive point charges of 12- C and 8- C are 10cm apart- The work done in bringing them 4cm closer is

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Potential Energy of a System of Multiple Point Charges

Two positive ...

Question

Two positive point charges of $12 μ C$ and $8 μ C$ are $10 c m$ apart. The work done in bringing them $4 c m$ closer is

5.8 J

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5.8 e V

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13 J

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13 e V

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Solution

The correct option is A $5.8 J$
$r_{1} = 10 c m = 0.1 m$

Final distance $= (10 - 4) c m = 6 c m = 0.06 m$
Work done, W $= 9 \times 10^{9} \times 10^{- 6} \times 8 \times 12 \times 10^{- 6} [\frac{1}{0.06} - \frac{1}{0.1}] = 5.76 J \approx 5.8 J$

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Two positive point charges of 12μC and 8μC are 10cm apart. The work done in bringing them 4cm closer is

The correct option is A 5.8Jr1=10cm=0.1mFinal distance=(10−4)cm=6cm=0.06mWork done, W =9×109×10−6×8×12×10−6[10.06−10.1]=5.76J≈5.8J

Two positive point charges of $12 μ C$ and $8 μ C$ are $10 c m$ apart. The work done in bringing them $4 c m$ closer is

The correct option is A $5.8 J$
$r_{1} = 10 c m = 0.1 m$

Final distance $= (10 - 4) c m = 6 c m = 0.06 m$
Work done, W $= 9 \times 10^{9} \times 10^{- 6} \times 8 \times 12 \times 10^{- 6} [\frac{1}{0.06} - \frac{1}{0.1}] = 5.76 J \approx 5.8 J$