voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]


=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v


charge on 2.0μf capacitor is


=5.32v
learn more about series capacitance from here: brainly.com/question/28166078
#SPJ4
In solids: All metals are good conductors of electricity as they contain free moving electrons. Non-metals doesn't conduct , but we consider Graphite the only non-metal that can conduct electricity for the presence of free moving electrons.
In Liquids ; Ionic compunds contains free moving ions , so they conduct electricity as well .
For this case, the first thing you should do is define a reference system.
Once the system is defined, we must follow the following steps:
1) Do the sum of forces in a horizontal direction
2) Do the sum of forces in vertical direction
The forces will be balanced if for each direction the net force is equal to zero.
The forces will be unbalanced if for each direction the net force is nonzero.
Answer:
Add the forces in the horizontal and vertical directions separately.
I believe the answer would be B) ...... sorry if i'm wrong