Question

The electric field strength in the capacitor shown in circuit below in steady state is $E=50 \, V / cm$. The distance between the plates of the capacitor $C$ is $0.5\, mm$, square plates are of area $100\, cm ^{2}$, the resistance $R=5\, \Omega$ and the internal resistance of battery is $r=0.1\, \Omega$.

• A. the emf of the battery is $2.55\, V$
• B. the attractive force between the plates is $2.2 \times 10^{-4} N$ (approx)
• C. the charge on the plates is $42.25 \times 10^{-10} C$
• D. the current through the battery in steady state is $0.5\, A$

Answer 1

Answer: (A), (D)

In steady state the potential difference across capacitor is given as
$V=E d=50 \times 0.05=2.5\, V$
Current through the battery in steady state is given as
$i=\frac{E}{r+R}=\frac{E}{0.1+5}=\frac{E}{5.1}\,\,\,\,...(1)$
Potential difference across capacitor is given as
$\frac{E}{5.1} \times 5 =2.5$
$\Rightarrow E =2.55\, V$
Current through battery is given by equation-(1) as $0.05\, A$.
Charge on capacitor plates is given as
$q=\epsilon_{0} A E=8.85 \times 10^{-12} \times 100 \times 10^{-4} \times 5000$
$\Rightarrow q=42.25 \times 10^{-11} C$
Force between plates of capacitor is given as
$F=\frac{1}{2} q E=0.5 \times 44.25 \times 10^{-11} \times 5000$
$=1.1 \times 10^{-6} N$