Question

6.2 What is the electric field strength 0.5 nm fr proton? SOLUTION \[ \begin{aligned} E & =\frac{Q}{4 \pi \varepsilon_{0} r^{2}} \\ & =\frac{1.602 \times 10^{-19} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)\left(0.5 \times 10^{-9} \mathrm{~m}\right)^{2}} \\ & =5.759 \times 10^{9} \mathrm{Vm}^{-1} \end{aligned} \]

          6.2 What is the electric field strength 0.5 nm fr proton?

SOLUTION
\[
\begin{aligned}
E & =\frac{Q}{4 \pi \varepsilon_{0} r^{2}} \\
& =\frac{1.602 \times 10^{-19} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)\left(0.5 \times 10^{-9} \mathrm{~m}\right)^{2}} \\
& =5.759 \times 10^{9} \mathrm{Vm}^{-1}
\end{aligned}
\]

6.2 What is the electric field strength 0.5 nm fr proton?

SOLUTION

E =(Q)/(4 πε0 r^2)
=(1.602 × 10^-19C)/(4 π(8.854 × 10^-12C^2 N^-1 m^-2)(0.5 × 10^-9 m)^2)
=5.759 × 10^9Vm^-1

Added by Margaret P.

Physical Chemistry

Robert J. Silbey, Robert A. Alberty,… 4th Edition

Chapter 7

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Solved by Expert Rachel Gore

Step 1

The electric field \( E \) due to a point charge is given by: \[ E = \frac{Q}{4 \pi \varepsilon_{0} r^{2}} \] Show more…

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6.2 What is the electric field strength 0.5 nm fr proton? SOLUTION \[ \begin{aligned} E & =\frac{Q}{4 \pi \varepsilon_{0} r^{2}} \\ & =\frac{1.602 \times 10^{-19} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)\left(0.5 \times 10^{-9} \mathrm{~m}\right)^{2}} \\ & =5.759 \times 10^{9} \mathrm{Vm}^{-1} \end{aligned} \]