1. A ciarge of $$ 2 \mu \mathrm{C} $$ is placed 0.1 m away from another charge of $$ -3 \mu \mathrm{C} $$. Using Coulomb's law. calculate the force between them. (Assume $$ \left.k=8.99 imes 10^{9} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}^{2} ight)(5 \mathrm{Marks} $$ ) Calculate the electric field at a point where a $$ 5 \mu \mathrm{C} $$ charge experiences a force of 0.2 N . ( 5 Marks) An 1 C voltage of $$ V(t)=220 \sin (100 \pi t) $$ is applied to a $$ 50 \Omega $$ resistor. Calculate the pecie current through the resistor. ( 5 Marks) A capacitor with a capacitance of $$ 20 \mu \mathrm{~F} $$ is connected in an AC circuit with a frequency of 50 Hz . Calculate the capacitive reactance. ( 5 Marks )

Question

1. A ciarge of $$ 2 \mu \mathrm{C} $$ is placed 0.1 m away from another charge of $$ -3 \mu \mathrm{C} $$. Using Coulomb's law. calculate the force between them. (Assume $$ \left.k=8.99 	imes 10^{9} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}^{2}ight)(5 \mathrm{Marks} $$ ) Calculate the electric field at a point where a $$ 5 \mu \mathrm{C} $$ charge experiences a force of 0.2 N . ( 5 Marks) An 1 C voltage of $$ V(t)=220 \sin (100 \pi t) $$ is applied to a $$ 50 \Omega $$ resistor. Calculate the pecie current through the resistor. ( 5 Marks) A capacitor with a capacitance of $$ 20 \mu \mathrm{~F} $$ is connected in an AC circuit with a frequency of 50 Hz . Calculate the capacitive reactance. ( 5 Marks )

UpStudy AI · Accepted Answer

Let's solve each part of your question step by step.

### Part 1: Force between the charges using Coulomb's Law

Coulomb's Law states that the force $$ F $$ between two point charges $$ q_1 $$ and $$ q_2 $$ separated by a distance $$ r $$ is given by:

$$
F = k \frac{|q_1 q_2|}{r^2}
$$

Where:
- $$ k = 8.99 \times 10^9 \, \mathrm{N \, m^2/C^2} $$
- $$ q_1 = 2 \, \mu \mathrm{C} = 2 \times 10^{-6} \, \mathrm{C} $$
- $$ q_2 = -3 \, \mu \mathrm{C} = -3 \times 10^{-6} \, \mathrm{C} $$
- $$ r = 0.1 \, \mathrm{m} $$

Now, substituting the values into the formula:

$$
F = 8.99 \times 10^9 \frac{|(2 \times 10^{-6})(-3 \times 10^{-6})|}{(0.1)^2}
$$

Calculating the absolute value of the product of the charges:

$$
|q_1 q_2| = |(2 \times 10^{-6})(-3 \times 10^{-6})| = 6 \times 10^{-12} \, \mathrm{C^2}
$$

Now substituting this into the force equation:

$$
F = 8.99 \times 10^9 \frac{6 \times 10^{-12}}{0.01}
$$

Calculating the denominator:

$$
0.01 = (0.1)^2
$$

Now substituting:

$$
F = 8.99 \times 10^9 \times 6 \times 10^{-10}
$$

Calculating:

$$
F = 53.94 \, \mathrm{N}
$$

Since the charges are of opposite signs, the force is attractive.

### Part 2: Electric field at a point where a $$ 5 \mu \mathrm{C} $$ charge experiences a force of 0.2 N

The electric field $$ E $$ at a point is given by the formula:

$$
E = \frac{F}{q}
$$

Where:
- $$ F = 0.2 \, \mathrm{N} $$
- $$ q = 5 \, \mu \mathrm{C} = 5 \times 10^{-6} \, \mathrm{C} $$

Substituting the values:

$$
E = \frac{0.2}{5 \times 10^{-6}}
$$

Calculating:

$$
E = \frac{0.2}{5 \times 10^{-6}} = 4 \times 10^4 \, \mathrm{N/C}
$$

### Part 3: Current through a resistor with AC voltage

The voltage is given by:

$$
V(t) = 220 \sin(100 \pi t)
$$

The RMS (Root Mean Square) voltage $$ V_{rms} $$ for a sinusoidal voltage is:

$$
V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{220}{\sqrt{2}} \approx 155.56 \, \mathrm{V}
$$

Using Ohm's Law, the current $$ I $$ through a resistor $$ R $$ is given by:

$$
I = \frac{V_{rms}}{R}
$$

Where $$ R = 50 \, \Omega $$:

$$
I = \frac{155.56}{50} \approx 3.11 \, \mathrm{A}
$$

### Part 4: Capacitive reactance of a capacitor in an AC circuit

The capacitive reactance $$ X_C $$ is given by:

$$
X_C = \frac{1}{2 \pi f C}
$$

Where:
- $$ f = 50 \, \mathrm{Hz} $$
- $$ C = 20 \, \mu \mathrm{F} = 20 \times 10^{-6} \, \mathrm{F} $$

Substituting the values:

$$
X_C = \frac{1}{2 \pi (50)(20 \times 10^{-6})}
$$

Calculating:

$$
X_C = \frac{1}{2 \pi (50)(20 \times 10^{-6})} = \frac{1}{6.2832 \times 0.001} \approx 159.15 \, \Omega
$$

### Summary of Results
1. Force between the charges: $$ 53.94 \, \mathrm{N} $$ (attractive)
2. Electric field: $$ 4 \times 10^4 \, \mathrm{N/C} $$
3. Current through the resistor: $$ 3.11 \, \mathrm{A} $$
4. Capacitive reactance: $$ 159.15 \, \Omega $$
1. Force between charges: 53.94 N (attractive) 2. Electric field: 4 x 10^4 N/C 3. Current through resistor: 3.11 A 4. Capacitive reactance: 159.15 Ω

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Physics Questions