Two charged particles, having equal charges of $2.0 \times 10^{- 5} C$ each , are brought from nfinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.

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Solution

$q_{1} - q_{2} = 2 \times 10^{- 5} C$
Each are brought from infinity to 10 cm apart d $= 10 \times 10^{- 2}$ m
So, work done = negative of work done (Potential E)
$= 9 \times 10^{9} \times \frac{2 \times 10^{- 5} \times 2 \times 10^{- 5}}{10 \times 10^{- 2}} = 36 J$

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Two charged particles, having equal charges of 2.0×10−5C each , are brought from nfinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.

q1−q2=2×10−5C Each are brought from infinity to 10 cm apart d =10×10−2m So, work done = negative of work done (Potential E) =9×109×2×10−5×2×10−510×10−2=36J

Two charged particles, having equal charges of $2.0 \times 10^{- 5} C$ each , are brought from nfinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.

$q_{1} - q_{2} = 2 \times 10^{- 5} C$
Each are brought from infinity to 10 cm apart d $= 10 \times 10^{- 2}$ m
So, work done = negative of work done (Potential E)
$= 9 \times 10^{9} \times \frac{2 \times 10^{- 5} \times 2 \times 10^{- 5}}{10 \times 10^{- 2}} = 36 J$