Chapter 24: Q4P (page 710)

Two large, parallel, conducting plates are 12 cmapart and have charges of equal magnitude and opposite sign on their facing surfaces. An electric force of 3.9 x 10^-15 Nacts on an electron placed anywhere between the two plates. (Neglect fringing.) (a) Find the electric field at the position of the electron. (b) What is the potential difference between the plates?

Short Answer

Expert verified

$E = 2.4 \times 10^{- 4} V / m$
$∆ V = 2.9 \times 10^{3} V$

Step by step solution

Given

The distance between two parallel, conducting plates is 12 cm.
The electric force is 3.9 x 10^-15 N.

Understanding the concept

Here we are using the equations of electric field energy $E = \frac{F}{q}$ and the potential difference V = Ed.

(a) Calculate the electric field at the position of the electron

$\begin{array}{rcl} E & = & \frac{F}{e} \\ = & \frac{(3.9 \times 10^{- 15} N)}{(1.6 \times 10^{- 19} C)} \\ = & 2.4 \times 10^{4} N / C \\ = & 2.4 \times 10^{4} V / m \end{array}$

Hence, the electric field at the position of the electron is $\begin{array}{rcl} E & = & 2.4 \times 10^{- 4} V / m \end{array}$

(b) Calculate the potential difference between the plates

$\begin{array}{rcl} ∆ V & = & E ∆ s \\ = & 2.4 \times 10^{4} V / m \times 0.12 m \\ = & 2.9 \times 10^{3} V \end{array}$

Hence, the potential difference between the plates is $∆ V \begin{array}{rcl} = \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 9 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} \end{array}$ ³V