Question

Three charges 2uc,3uc,5uc are located on a straight line as shown in the figure.
Find :- i) P.E of 5uc charge
ii) P.E of system of three charges



please answer aptly to this question
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Answer 1

i) PE(5uC) = PE(5,3) + PE(5,2)

= k(5u)(3u)/1 + k(5u)(2u)/2

= k(5u)[3u + 2u/2]

= k(5u)[3u + 1u]

= k(5u)(4u)

= 9×10^9(5×10^-6)(4×10^-6)

= 180 × 10^-3

= 0.18 Joules

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ii) PE of the system = PE(5,2) + PE(5,3) + PE(2,3)

And we have calculated above that PE(5,2) + PE(5,3) = 0.18 Joules.

Thus, PE(5,2) + PE(5,3) + PE(2,3) = 0.18 + k(2u)(3u)/2

= 0.18 + k(1u)(3u)

= 0.18 + 9×10^9(10^-6)(3×10^-6)

= 0.18 + 27×10-3

= 0.18 + 0.027

= 0.207 Joules.