1) The plates of a parallel plate capacitor 5x10^-3 m apart are maintained at a potential difference of 5x10^4 V. Calculate the magnitude of electric field intensity between the pl... 1) The plates of a parallel plate capacitor 5x10^-3 m apart are maintained at a potential difference of 5x10^4 V. Calculate the magnitude of electric field intensity between the plates (ii) force on the electron (iii) acceleration of the electron (electron = 1.6x10^-19 C, mass of electron = 9.1x10^-31 kg). Food and Nutrition: Make a food menu card for a pregnant/lactating mother from Monday to Sunday. Ensure varieties and that the meals are balanced.
Understand the Problem
The question is asking to calculate specific physics-related values involving a capacitor and the forces acting on an electron, as well as to create a balanced food menu for a specific dietary need.
Answer
$E = 1 \times 10^7 \, \text{V/m}, \, F = 1.6 \times 10^{-12} \, \text{N}, \, a \approx 1.76 \times 10^{18} \, \text{m/s}^2$
Answer for screen readers
The magnitude of electric field intensity is $E = 1 \times 10^7 , \text{V/m}$, the force on the electron is $F = 1.6 \times 10^{-12} , \text{N}$, and the acceleration of the electron is approximately $a \approx 1.76 \times 10^{18} , \text{m/s}^2$.
Steps to Solve
- Calculate Electric Field Intensity
The electric field intensity ($E$) between the plates of a capacitor can be calculated using the formula:
$$ E = \frac{V}{d} $$
Where:
- $V$ is the potential difference (5 x $10^4$ V)
- $d$ is the distance between the plates (5 x $10^{-3}$ m)
Substituting the values:
$$ E = \frac{5 \times 10^4 , \text{V}}{5 \times 10^{-3} , \text{m}} = 1 \times 10^7 , \text{V/m} $$
- Calculate the Force on the Electron
The force ($F$) on an electron in an electric field is given by the formula:
$$ F = qE $$
Where:
- $q$ is the charge of an electron (1.6 x $10^{-19}$ C)
- $E$ is the electric field intensity (calculated in the previous step)
Substituting the values:
$$ F = (1.6 \times 10^{-19} , \text{C}) \times (1 \times 10^7 , \text{V/m}) = 1.6 \times 10^{-12} , \text{N} $$
- Calculate the Acceleration of the Electron
The acceleration ($a$) of the electron can be calculated using Newton's second law:
$$ a = \frac{F}{m} $$
Where:
- $F$ is the force calculated earlier (1.6 x $10^{-12}$ N)
- $m$ is the mass of the electron (9.1 x $10^{-31}$ kg)
Substituting the values:
$$ a = \frac{1.6 \times 10^{-12} , \text{N}}{9.1 \times 10^{-31} , \text{kg}} \approx 1.76 \times 10^{18} , \text{m/s}^2 $$
The magnitude of electric field intensity is $E = 1 \times 10^7 , \text{V/m}$, the force on the electron is $F = 1.6 \times 10^{-12} , \text{N}$, and the acceleration of the electron is approximately $a \approx 1.76 \times 10^{18} , \text{m/s}^2$.
More Information
This problem involves concepts from electric fields and forces acting on charged particles. The calculations emphasize the relationships between voltage, distance, force, and acceleration in an electric context.
Tips
- Forgetting to convert units properly can lead to incorrect results. Always ensure that the units are consistent throughout the calculations.
- Mixing up the order of operations in calculations, especially with division and multiplication, can lead to errors.
AI-generated content may contain errors. Please verify critical information
