Answer:
a.[tex]72.8\mu C[/tex]
b.[tex]72.8\mu C[/tex]
Explanation:
We are given that
Capacitance of one capacitor=[tex]2.6\mu F[/tex]
Capacitance of second capacitor=[tex]2.6\mu F[/tex]
Potential difference of battery=28 v
a.When one of the capacitor is squeezed so that plate separation is halved then we have to find the amount of additional charge is transferred to the capacitor by the battery.
Initial capacitance, [tex]C=\frac{\epsilon_0A}{d}[/tex]
If the separation is halved , then the capacitance will be doubled
[tex]q=CV[/tex]
If the capacitance will be doubled then the charge will be doubled also.
Initial charge=[tex]2.6\times 10^{-6}\times 28=72.8\mu C[/tex]
Final charge on the capacitor
[tex]q=2CV=2\times 72.8\mu C=145.6\mu C[/tex]
Additional charge transmitted, [tex]q'=145.6-72.8=72.8\mu C[/tex]
b.We have to find the increase in the total charge stored on the capacitors
Initial total charge is given by
[tex]q=(c_1+c_2)V=(2.6+2.6)\times 28=145.6 \mu C[/tex]
Final total charge on the capacitor
[tex]q=(2\times 2.6+2.6)\times 28=218.4\muC[/tex]
Increase in the charge=[tex]q'-q=218.4-145.6=72.8\mu C[/tex]
