Question

$FIGUREQ25.10$ shows two points near a positive point charge.
$\left(a\right)$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

$\left(a\right)$The ratio of electric potentials is $V_2/V_1=1/3.$

$\left(b\right)$The ratio of electric field strength is $E_2/E_1=1/9.$

Step-by-step solution

Step: 1 Electric potential:

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 conductor to the voltage differential between them is known as capacitance.

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

Charge $q$is same.

$V=\frac{q}{4\pi {ϵ}_{0}r}$

role="math" localid="1648811378097" $$\begin{gathered} V_2/V_1=r_1/r_2 \\ {V}_2/V_1=1/3. \end{gathered}$$

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

The charge $q$is contant,

$$\begin{gathered} E=\frac{q}{4\pi {ϵ}_0{r}^2} \\ {E}_2/E_1=r_1^2/r_2^2 \\ {E}_2/E_1=1/9. \end{gathered}$$