Respuesta :
Answer:
Explanation:
In the first case,
Q₀ = C₁ V, where C₁ is equivalent capacitance of two capacitor having capacitance C each.
C₁ = C / 2
Q₀ = CV / 2
CV = 2Q₀
Total charge = 2Q₀ = CV
In the second case ,
capacitance will be kC and C .
equivalent capacitance
C₂ = kC x C / ( kC + C)
= kC / ( 1+k)
Charge on each capacitor = V x C₂
= kCV / ( 1+k)
= 2kQ₀ / ( 1+k)
This will be the charge on each capacitor. So
total charge = 4kQ₀ / ( 1+k)
The total positive charge stored on the two capacitors is;
4kQ₀/(k + 1)
We are told that the charge on each capacitor is Q₀.
Now, formula for charge stored on a capacitor is;
Q = CV
Where;
Q is charge
C is capacitance
V is voltage
Thus, total charge stored on the two capacitors will be;
Q = Q₁ + Q₂
Since Q₀ is the charge on each capacitor, then;
For Capacitor 1, we have; Q₀ = C₁V
For capacitor 2, we have; Q₀ = C₂V
Thus, let the capacitance C₁ be be half the capacitance C of two capacitors.
Thus;
C₁ = ¹/₂C
Thus;
Q₀ = ¹/₂CV
2Q₀ = CV
Now, a dielectric with dielectric constant k > 1 is inserted between the plates of capacitor C₁. Thus, the two capacitors will now be kC and C.
Thus;
C₂ = (kC × C)/(kC + C)
C will cancel out to give;
C₂ = kC/(k + 1)
Earlier on, we saw that;
Q₀ = C₂V
Thus;
Q₀ = kCV/(k + 1)
Earlier, we saw that 2Q₀ = CV
Thus, charge is;
Q₀ = 2kQ₀/(k + 1)
Since 2 capacitors, then total positive charge stored is;
2 × 2kQ₀/(k + 1) = 4kQ₀/(k + 1)
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