Question

The signal from the oscillating electrode is fed into an amplifier, which reports the measured voltage as an rms value, 1.0 nV . What is the potential difference between the two extremes

Answer 1

Answer:

The value is  $V = 2.8284 *10^{-9 } \ Volts$

Explanation:

From the question we are told that

   The measure voltage is  $E_{rms} = 1.0 \ n V = 1.0 *10^{-9} \ V$

   

Generally the peak voltage  is   mathematically represented as

       $E_{max} = \sqrt{2} * E_{rms}$

=>    $E_{max} = \sqrt{2} * 1.0 *10^{-9}$

=>    $E_{max} = 1.4142 *10^{-9} \ volts$

Generally the potential difference between the two extremes is mathematically represented as

       $V = 2 * E_{max}$

=>    $V = 2 * 1.4142 *10^{-9}$

=>    $V = 2.8284 *10^{-9 } \ Volts$