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:
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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 \(\( 10^3 \, \text{V} \)\) requires \(\( 2 \times 10^{-3} \, \text{J} \)\) of work. What is the magnitude of the charge?
To determine the magnitude of the charge, \(\( q \)\), we can use the relationship between work (W), charge (q), and potential difference (V):
\(\[ W = q \cdot V \]\)
Rearranging to solve for \(\( q \)\):
\(\[ q = \frac{W}{V} \]\)
Given:
\(\[ W = 2 \times 10^{-3} \, \text{J} \]\)
\(\[ V = 10^3 \, \text{V} \]\)
Substituting the values:
\(\[ q = \frac{2 \times 10^{-3}}{10^3} = 2 \times 10^{-6} \, \text{C} \]\)
Therefore, the magnitude of the charge is \(\( 2 \times 10^{-6} \, \text{C} \)\).
10. Calculate the potential difference between two points which are 20 cm and 30 cm from an isolated point charge of \(\( 3 \times 10^{-8} \, \text{C} \)\) in a vacuum.
To determine the potential difference between two points at distances \(\( r_1 \)\) and \(\( r_2 \)\) from a point charge \(\( q \)\), we use the formula for the electric potential \(\( V \)\) at a distance \(\( r \)\) from a point charge:
\(\[ V = \frac{k \cdot q}{r} \]\)
Where:
- \(\( k \)\) is Coulomb's constant, approximately \(\( 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)\)
The potential difference \(\( \Delta V \)\) between the two points is:
\(\[ \Delta V = V_1 - V_2 = \frac{kq}{r_1} - \frac{kq}{r_2} \]\)
Given:
\(\[ q = 3 \times 10^{-8} \, \text{C} \]\)
\(\[ r_1 = 20 \, \text{cm} = 0.2 \, \text{m} \]\)
\(\[ r_2 = 30 \, \text{cm} = 0.3 \, \text{m} \]\)
Substituting these values:
\(\[ \Delta V = \frac{9 \times 10^9 \times 3 \times 10^{-8}}{0.2} - \frac{9 \times 10^9 \times 3 \times 10^{-8}}{0.3} \]\)
\(\[ \Delta V = \frac{27 \times 10^1}{0.2} - \frac{27 \times 10^1}{0.3} \]\)
\(\[ \Delta V = 1350 \, \text{V} - 900 \, \text{V} \]\)
\(\[ \Delta V = 450 \, \text{V} \]\)
Therefore, the potential difference between the two points is \(\( 450 \, \text{V} \)\).
