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
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Answer with Explanation:
We are given that
a.Length of segment,l=20 m
Magnetic force ,F=
Substitute the values
Hence, the magnetic force exert by each segment on the other=0.0119 N
b.We know that when current carrying in the wires are in same direction then the force will attract to each other.
Hence, the force will be attractive.
Am not really sure but what i see is D
The best and most correct answer among the choices provided by your question is the second choice.
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