Question

$FIGUREQ25.9$shows two points inside a capacitor. Let$V=0V$ at the negative plate.
$\left(a\right)$What is the ratio$\raisebox{1ex}{$V_2$}\!\left/ \!\raisebox{-1ex}{$V_1$}\right.$ of the electric potentials? Explain.
$\left(b\right)$What is the ratio$\raisebox{1ex}{$E_2$}\!\left/ \!\raisebox{-1ex}{$E_1$}\right.$of the electric field strengths?

Short answer

Part $\left(a\right)$

$\left(a\right)$The ratio of eletric potentials is $V_2/V_1=E{s}_2/E{s}_1=3.$

Part$\left(b\right)$

$\left(b\right)$The ratio of eletric field strength is$E_2/E_1=1.$

Step-by-step solution

Step: 1 Capacitor at negative plate:

Electrical capacitance is measured in Farads and is represented as a ratio of charge to potential difference. The electric charge on each conductor is directly proportional to the potential between them, and this is how capacitance is calculated.

Step: 2 Ratio of electric potential: (part a)

Electrical capacitance is measured in Farads and is defined as the ratio of charge to potential difference. The ratio of the electromotive force on each circuit to the potential difference between them is known as capacitance.$V_2/V_1=E{s}_2/E{s}_1=3.$

Step: 3 Ratio of electric field strength: (part b)

The parallel plate capacitor is constant inside field,

$$\begin{gathered} E=\frac{\eta }{{ϵ}_0}, \\ {E}_2/E_1=1. \end{gathered}$$