Certainly! Let's address each question step-by-step.
7. Where do we find capacitors in real life?
Capacitors are found in a wide range of real-life applications, including:
- Electronic devices like smartphones, laptops, and televisions
- Power supply circuits to smooth out voltage fluctuations
- Radio frequency circuits, such as those in tuning radios
- Electric motors and air conditioners to improve efficiency and power factor correction
- Camera flashes to quickly release stored energy
- Energy storage in applications like uninterruptible power supplies (UPS)
8. What is the symbol of capacitors?
The symbol for a capacitor consists of two parallel lines, one representing each plate of the capacitor. It looks like this:
```
| |
```
In the context of schematic diagrams, it can also be represented with a "+" and "-" sign if it is a polarized capacitor, such as an electrolytic capacitor.
9. To move a point charge through a potential difference of [tex]\( 10^3 \, \text{V} \)[/tex] requires [tex]\( 2 \times 10^{-3} \, \text{J} \)[/tex] of work. What is the magnitude of the charge?
To determine the magnitude of the charge, [tex]\( q \)[/tex], we can use the relationship between work (W), charge (q), and potential difference (V):
[tex]\[ W = q \cdot V \][/tex]
Rearranging to solve for [tex]\( q \)[/tex]:
[tex]\[ q = \frac{W}{V} \][/tex]
Given:
[tex]\[ W = 2 \times 10^{-3} \, \text{J} \][/tex]
[tex]\[ V = 10^3 \, \text{V} \][/tex]
Substituting the values:
[tex]\[ q = \frac{2 \times 10^{-3}}{10^3} = 2 \times 10^{-6} \, \text{C} \][/tex]
Therefore, the magnitude of the charge is [tex]\( 2 \times 10^{-6} \, \text{C} \)[/tex].
10. Calculate the potential difference between two points which are 20 cm and 30 cm from an isolated point charge of [tex]\( 3 \times 10^{-8} \, \text{C} \)[/tex] in a vacuum.
To determine the potential difference between two points at distances [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] from a point charge [tex]\( q \)[/tex], we use the formula for the electric potential [tex]\( V \)[/tex] at a distance [tex]\( r \)[/tex] from a point charge:
[tex]\[ V = \frac{k \cdot q}{r} \][/tex]
Where:
- [tex]\( k \)[/tex] is Coulomb's constant, approximately [tex]\( 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)[/tex]
The potential difference [tex]\( \Delta V \)[/tex] between the two points is:
[tex]\[ \Delta V = V_1 - V_2 = \frac{kq}{r_1} - \frac{kq}{r_2} \][/tex]
Given:
[tex]\[ q = 3 \times 10^{-8} \, \text{C} \][/tex]
[tex]\[ r_1 = 20 \, \text{cm} = 0.2 \, \text{m} \][/tex]
[tex]\[ r_2 = 30 \, \text{cm} = 0.3 \, \text{m} \][/tex]
Substituting these values:
[tex]\[ \Delta V = \frac{9 \times 10^9 \times 3 \times 10^{-8}}{0.2} - \frac{9 \times 10^9 \times 3 \times 10^{-8}}{0.3} \][/tex]
[tex]\[ \Delta V = \frac{27 \times 10^1}{0.2} - \frac{27 \times 10^1}{0.3} \][/tex]
[tex]\[ \Delta V = 1350 \, \text{V} - 900 \, \text{V} \][/tex]
[tex]\[ \Delta V = 450 \, \text{V} \][/tex]
Therefore, the potential difference between the two points is [tex]\( 450 \, \text{V} \)[/tex].
