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Expert-verified Found in: Page 696 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # A doubly charged ion is accelerated to an energy of $$32.0{\rm{ }}keV$$ by the electric field between two parallel conducting plates separated by $$2.00{\rm{ }}cm$$. What is the electric field strength between the plates?

Value of the electric strength between the plates is $$E = 8.00 \times {10^5}\;V/m$$.

See the step by step solution

## Step 1: Definition of electric volt and electric potential energy.

Electron Volt: is the energy given to a fundamental charge accelerated through a potential difference of $$1\;V$$. In equation form,

$$1{\rm{ }}eV = \left( {1.60 \times {{10}^{ - 19}}{\rm{ }}C} \right)(1\;V) = 1.60 \times {10^{ - 19}}\;J$$

Electric Potential Energy: If a particle with charge $$q$$ is positioned at a location where a charged object's electric potential is $$V$$, the particle-object system's electric potential energy is $${U_e} = qV$$.

And the potential difference between two points separated by a distance $$d$$ in a uniform electric field of magnitude $$E$$ is $$\Delta V = Ed$$.

## Step 2: Given data for the calculation of electric field strength.

As, the ion is doubly charged.

The electric potential energy of the ion is:

$$\begin{array}{c}{U_e} = (32.0{\rm{ }}keV)\left( {\frac{{1000{\rm{ }}eV}}{{1{\rm{ }}keV}}} \right)\\{U_e} = 32.0 \times {10^3}{\rm{ }}eV\end{array}$$

The distance between the two plates is:

$$d = (2.00\;cm)\left( {\frac{{1\;m}}{{100\;cm}}} \right)$$

The value of the fundamental charge is:

$$e = 1.60 \times {10^{ - 19}}{\rm{ }}C$$

## Step 3: Converting electric potential energy from electron volt to Joule.

Convert the electric potential energy of the ion from electron volts to joules using the conversion factor in equation below:

$$\begin{array}{c}{U_e} = \left( {32.0 \times {{10}^3}{\rm{ }}eV} \right)\left( {\frac{{1.60 \times {{10}^{ - 19}}\;J}}{{1{\rm{ }}eV}}} \right)\\{U_e} = 5.12 \times {10^{ - 15}}\;J\end{array}$$

Hence, the potential energy from electron volt to joule is $$5.12 \times {10^{ - 15}}\;J$$.

## Step 4: Calculation for the potential difference between the plates.

The potential difference between the two plates is found by solving equation below for $$\Delta V$$:

$$\Delta V = \frac{{{U_e}}}{q}$$

Where $$q = 2e$$ is the charge of the ion:

$$\Delta V = \frac{{{U_e}}}{{2e}}$$

Entering the values for $${U_e}$$ and $$e$$, we obtain:

$$\begin{array}{c}\Delta V = \frac{{5.12 \times {{10}^{ - 15}}\;J}}{{2\left( {1.60 \times {{10}^{ - 19}}C} \right)}}\\\Delta V = 1.60 \times {10^4}\;V\end{array}$$

Hence, the potential difference between the plates is $$1.60 \times {10^4}\;V$$ .

## Step 5: Calculation for the electric field strength.

The electric field strength between the two plates is found by solving equation below for $$E$$:

$$E = \frac{{\Delta V}}{d}$$

Substitute numerical values:

$$\begin{array}{c}E = \frac{{1.60 \times {{10}^4}\;V}}{{0.0200\;m}}\\E = 8.00 \times {10^5}\;V/m\end{array}$$

Therefore, electric field value is obtained as $$8.00 \times {10^5}\;V/m$$. ### Want to see more solutions like these? 