Question

A parallel plate capacitor of capacitance $800\mu$F and separation 1cm is charged to a potential difference $100V$. The battery is then disconnected. Electromagnetic waves are incident on a negatively charged plate that emit electrons with kinetic energy ranging from $0$ to $1.5eV.$ The electrons are attracted to a positive plate and a constant current flows between the plates whose value is ${10}^{-12}$ A for time t in seconds. If $t={10}^{10}y$. Then the value of y is :

• A. $8$
• B. $4$
• C. $12$
• D. $16$

Answer 1

Answer: (A)

$8$

The capacitance $C$ is given by,
$C=\frac{Q}{V}$, where $Q$ is the charge and $V$ is the potential difference.
Also, Charge, $Q=It$, where $I$ is the current and $t$ is the time.
Hence we have,
$C=\frac{I\times t}{V}$
$⟹t=\frac{C\times V}{I}$
i.e, $t=\frac{800\times {10}^{-6}\times 100}{{10}^{-12}}=8\times {10}^{10}sec$
Hence $y=8$