Question

Answer this please. First answer will be marked Brainliest. ​

Answer 1

Equivalent capacitance in the above branch will be  

4+9+3

4(9+3)

​  

μF=3μF.  

Total charge in above branch will be Q=CV=24μC. This charge resides on the 4μF capacitor and 12μF (combination of 3 μF  and 9μF) capacitor.  

Now, voltage across 12μF combination of capacitors is given by V=Q/C=24/12=2V. This is the same as the voltage across 9μF capacitor.  

Hence, charge on 9μF capacitor is Q=CV=9×2=18μC

From above total charge on 4μF and 9μF capacitors is 24+18=42μC

Now, by coulomb's  law,

E=  

r  

2

 

kQ

​  

 

E=  

30  

2

 

9×10  

9

×42×10  

−6

 

​  

=420N/C