Question

please answer (b) part.

Answer 1

Answer is 2.304 * 10^9 joules.
Kindly refer to the attachment, Chandan!
Hope it will assist you! :-)


{ Posted two attachments of the same answer so that there's no obstruction in understanding any of the steps. :-) }

Answer 2

You need to understand what dissociating the system of charge means here.

if they the system of charge is dissociated, there will be no interaction among the charges. or they will be at infinite distance form each other which means the potential energy of the system will be 0.

So find the PE now and give the same amount of energy as work done so that PE becomes 0.
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distance between the charges is same for all.

r = 10cm = 0.1 m

PE of the system of charges=

PE = PE(q,-q) + PE(q,2q) + PE(2q,-q)

$PE = \frac{1}{4\pi \epsilon_{o} } ( \frac{q \times ( - q)}{0.1} ) + \frac{1}{4\pi \epsilon_{o} } ( \frac{q \times 2q}{0.1} ) + \frac{1}{4\pi \epsilon_{o} } ( \frac{2q \times ( - q)}{0.1} )$

$PE = \frac{1}{4\pi \epsilon_{o} } ( \frac{q \times ( - q)}{0.1} + \frac{q \times 2q}{0.1} + \frac{2q \times ( - q)}{0.1} ) \\ \\ PE = \frac{1}{4\pi \epsilon_{o} } ( \frac{ { - q}^{2} }{0.1} + \frac{2 {q}^{2} }{0.1} + \frac{ - 2 {q}^{2} }{0.1} ) \\ \\ PE = \frac{1}{4\pi \epsilon_{o} } ( \frac{ { - q}^{2} }{0.1}) \\ \\ PE = \frac{ - 10 { q }^{2} }{4\pi \epsilon_{o} }$

So potential of the system is $\frac{ - 10 {q}^{2} }{4\pi \epsilon_{o} }$

To make the potential energy of system 0, you need to provide $\frac{ + 10 {q}^{2} }{4\pi \epsilon_{o} }$ energy to the system.

So work done is $\frac{ + 10 {q}^{2} }{4\pi \epsilon_{o} }$