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 \(22.45 \mathrm{~m}\)