Find the potential differnce between the points A and B and that between E and F of the circuit in.
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Answer

`5V,-5V`

Explanation

The distribution of charge is shown in in compliance with the point rule. Apply loop rules.
Left loop: `-q_(1)/5+q_(3)/(0.75)+q_(1)/(15)=0`
or `q_(1)-3q_2+20 q_3=0` (i)
Middle loop : `-(q_(2)+q_(3))/(15)-q_(3)/(0.75)+(q_(1)-q_()3)/5-q_(3)/(0.75)=0`
or `3q_(1)-q_(2)-44q_(3)=0` (ii)
Outermost loop: `23-q_(2)/5-(q_(2)+q_(3))/(15)-q_(2)/5=0`
Outermost loop : `23-q_(2)/5-(q_(2)+q_(3))/(15)-q_(2)/5=0`
or `345=7q_(2)+q_(3)` (iii) `Solving for `q_(1),q_(2)`, and `q_(3)`. we get
`q_(1)=(19xx345)/(92),q_(2)=(13xx345)/(92),q_(3)=(345)/(92)`
Therefore, potential difference between `A`and `B` is
`q_(3)/(0.75)=(345)/(92)xx4/3=5 V`
Potential difference between `E` and `F` is also `5 V` but in the opposite direction.
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