Question

A +7.50 μC point charge is sitting at the origin.

What is the radial distance between the 500 V equipotential surface and the 1000 V surface?

What is the distance between the 1000 V surface and the 1500 V surface?

 

Answer 1

Given data. Point charge, $q=+7.50 \mu \mathrm{C}$

(a) Write the expression for the electrical potential $(V)$ created by a point charge at a

distance $r$

$$V=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$$

Here, $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N}-\mathrm{m}^{2}$ is the permittivity of free space.

Apply the above formula for $500 \mathrm{~V}$ equipotential surface.

$$500=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{1}}$$

Here, $r_{1}$ is the radial distance of $500 \mathrm{~V}$ equipotential surface from origin. Substitute the values in the above equation.

$$\begin{aligned} 500 &=\frac{1}{4 \pi\left(8.85 \times 10^{-12}\right)} \times \frac{7.5 \times 10^{-6}}{r_{1}} \\ 500 \times r_{1} &=8.98 \times 10^{9} \times 7.5 \times 10^{-6} \\ r_{1} &=134.7 \mathrm{~m} \end{aligned}$$

Apply formula (1) for $1000 \mathrm{~V}$ equipotential surface.

Apply formula (1) for $1000 \mathrm{~V}$ equipotential surface. $1000=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{2}}$

Here, $r_{2}$ is the radial distance of $1000 \mathrm{~V}$ equipotential surface from origin. Substitute the values in the above equation.

$$\begin{aligned} 1000 &=\frac{1}{4 \pi \times\left(8.85 \times 10^{-12}\right)} \times \frac{7.5 \times 10^{-6}}{r_{2}} \\ 1000 \times r_{2} &=8.98 \times 10^{9} \times 7.5 \times 10^{-6} \\ r_{2} &=67.35 \mathrm{~m} \end{aligned}$$

Thus, the radial distance $\left(r_{d}\right)$ between $500 \mathrm{~V}$ and $1000 \mathrm{~V}$ will be,

$r_{d}=r_{1}-r_{2}$

Substitute the value of $r_{1}$ and $r_{2}$.

$$\begin{aligned} r_{d} &=134.7-67.35 \\ &=67.35 \mathrm{~m} \end{aligned}$$

Hence, the radial distance between $500 \mathrm{~V}$ and $1000 \mathrm{~V}$ is $67.35 \mathrm{~m}$

(b)

Apply the equation (1) for $1500 \mathrm{~V}$ equipotential surface.

$$1500=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{3}}$$

Here, $r_{3}$ is the distance of $1500 \mathrm{~V}$ equipotential surface from origin.

Substitute the values in the above equation.

$$\begin{aligned} 1500 &=\frac{1}{4 \pi\left(8.85 \times 10^{-12}\right)} \frac{7.5 \times 10^{-6}}{r_{3}} \\ 1500 \times r_{3} &=8.98 \times 10^{9} \times 7.5 \times 10^{-6} \\ r_{3} &=44.9 \mathrm{~m} \end{aligned}$$

Thus, the radial distance $\left(r_{d}{ }^{\prime}\right)$ between $1000 \mathrm{~V}$ and $1500 \mathrm{~V}$ will be,

$$\begin{aligned} r_{d}{ }^{\prime} &=r_{2}-r_{3} \\ &=67.35-44.9 \\ &=22.45 \mathrm{~m} \end{aligned}$$

Hence, the distance between $1000 \mathrm{~V}$ and $1500 \mathrm{~V}$ is $22.45 \mathrm{~m}$

a) the radial distance between $500 \mathrm{~V}$ and $1000 \mathrm{~V}$ is $67.35 \mathrm{~m}$

b) the distance between $1000 \mathrm{~V}$ and $1500 \mathrm{~V}$ is $22.45 \mathrm{~m}$