Question

The deflection plates in an oscilloscope are 10cm by 2cm with a gap distance of 1mm. A 100V potential difference is suddenly applied to the initially uncharged plates through a $\mathbf{1000}\mathit{\Omega }$resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95V?

Short answer

$t=5.3\times {10}^{-8}s$

Step-by-step solution

Given data

Plate area $A=\left(10cm\times 2cm\right)$

Gap $s=1mm$

Potential difference $V=95V$

Max. Potential difference $V_{\max }=100V$

Resistance $R=1000\Omega$

Formula used

$\mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\left(1-exp\left(\frac{-t}{RC}\right)\right)$

Where,$V_{max}$is max. Potential difference reaches

$t$ Is time require

$R$ is resistance and$C$ is the capacitance

Calculation for time required

$\begin{array}{c}C=\frac{{\epsilon }_{0}A}{s}\\ =\frac{8.85\times {10}^{-12}\times 2\times {10}^{-3}}{1\times {10}^{-3}}\\ =1.77\times {10}^{-11}F\end{array}$

$\begin{array}{c}V=V_{max}\left(1-exp\left(\frac{-t}{RC}\right)\right)\\ 95=100\left(1-exp\left(\frac{-t}{1000\times 1.77\times {10}^{-11}}\right)\right)\\ 0.95=1-exp\left(\frac{-t}{1.77\times {10}^{-8}}\right)\\ exp\left(\frac{-t}{1.77\times {10}^{-8}}\right)=0.05\end{array}$

$t=5.3\times {10}^{-8}s$