Physics - Electrostatics Question with Solution | TestHub

$\therefore V_1=\left(\frac{\frac{15}{4}}{1+\frac{15}{4}}\right)\times 190$

$=\frac{15}{19}\times 190=150\mathrm{ V}$

$\Rightarrow V_2=\left(190-150\right)\mathrm{ V}=40\mathrm{ V}$

Now,



So, the potential difference between$A$and$B$will be,

$V_{AB}=\left(\frac{1}{1+3}\right)\times 40$

$=\frac{1}{4}\times 40=10\mathrm{ V}$