Question

The drawing shows the electrical potential as a function

Answer 1

Answer:

$a)\quad E_{ab} = 0 \quad Volts/m\\\\b)\quad E_{bc} = 10 \quad Volts/m\\\\c)\quad E_{cd} = 5 \quad Volts/m$

Explanation:

As figure is not given so considering the most relevant diagram for question attached below

We can define Electric field as rate of change of electric potential with respect to space (say x-axis here). It is related in formula as

$E=-\frac{\Delta V}{S}---(1)$

a) For Region A to B:

As can be seen from figure, point charge moves from 0 to 0.2 along x-axis, the value of electric potential remains constant i.e 5 volts

$E_{ab}=-\frac{V_{b}-V_{a}}{b-a}\\\\E_{ab}=-\frac{5-5}{0.2-0}\\\\E_{ab}=0\quad Volts/m$

b) For Region B to C:

As point charge moves from B to C (0.2 to 0.4) along x-axis, the value of electric potential decreases from 5 volts to 3 volts. Electric field induced is:

$E_{bc}=-\frac{V_{c}-V_{b}}{c-b}\\\\E_{bc}=-\frac{3-5}{0.4-0.2}\\\\E_{bc}=10\quad Volts/m$

c) For Region C to D:

As point charge moves from C to D (0.4 to 0.8) along x-axis, the value of electric potential decreases from 3 volts to 1 volt. Electric field induced is:

$E_{cd}=-\frac{V_{d}-V_{c}}{d-c}\\\\E_{cd}=-\frac{1-3}{0.8-0.4}\\\\E_{cd}=5\quad Volts/m$

For Information:

Unit of Electric field is V/m or N/C