Find the potential difference (inV) between points A and B shown in the circuit.

Explanation

The distribution of charge is shown in figure `(-q_2)/5+q_3/0.75+q_1/15=0`
`rArr q_1-3q_2+20q_3=0`…(i)
`-((q_2+q_3)/15)-q_3/0.75+(q_1-q_3)/5=q_3/0.75=0`
`rArr 3q_1-q_2-44q_3=0`...(ii)
`23-q_2/5-((q_2+q_3)/15)-q_2/5=0`
`345=7q_2+q_3`...(ii) From eq.(i),(ii),(iii)
`q_1=(19xx345)/92,q_2=(13xx345)/92,q_3=345/92`
Potential difference between A and B =`q_3/0.75=5V`