Class 10 - Unit # 14 : Electrostatics - Solved Numericals

1. What is the electric force of repulsion between two electrons at a distance of $1\text{ m}$?

Data:

• Charge on electron 1 ($q_1$) = $1.6 \times 10^{-19}\text{ C}$
• Charge on electron 2 ($q_2$) = $1.6 \times 10^{-19}\text{ C}$
• Distance ($r$) = $1\text{ m}$
• Coulomb's constant ($k$) = $9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$
• Electric force ($F$) = ?

Solution:

According to Coulomb's Law:

$$F = k \cdot \frac{q_1 \cdot q_2}{r^2}$$ $$F = (9 \times 10^9) \times \frac{(1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{(1)^2}$$ $$F = 9 \times 10^9 \times 2.56 \times 10^{-38}$$ $$F = 23.04 \times 10^{-29} = 2.304 \times 10^{-28}\text{ N}$$

RESULT: The electric force of repulsion between the electrons is $2.304 \times 10^{-28}\text{ N}$.

2. Two point charges $q_1 = 2\text{ }\mu\text{C}$ and $q_2 = 3\text{ }\mu\text{C}$ are placed at a distance of $5\text{ cm}$. What will be the Coulomb's force between them?

Data:

• Charge ($q_1$) = $2\text{ }\mu\text{C} = 2 \times 10^{-6}\text{ C}$
• Charge ($q_2$) = $3\text{ }\mu\text{C} = 3 \times 10^{-6}\text{ C}$
• Distance ($r$) = $5\text{ cm} = 0.05\text{ m}$
• Coulomb's constant ($k$) = $9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$
• Coulomb's force ($F$) = ?

Solution:

According to Coulomb's Law:

$$F = k \cdot \frac{q_1 \cdot q_2}{r^2}$$ $$F = (9 \times 10^9) \times \frac{(2 \times 10^{-6}) \times (3 \times 10^{-6})}{(0.05)^2}$$ $$F = \frac{9 \times 10^9 \times 6 \times 10^{-12}}{0.0025}$$ $$F = \frac{0.054}{0.0025} = 21.6\text{ N}$$

RESULT: The electric force between the charges is $21.6\text{ N}$.

3. A $2\text{ }\mu\text{C}$ charge is placed in an electric field of $3.42 \times 10^{11}\text{ N/C}$. What will be the force acting on it?

Data:

• Charge ($q$) = $2\text{ }\mu\text{C} = 2 \times 10^{-6}\text{ C}$
• Electric field intensity ($E$) = $3.42 \times 10^{11}\text{ N/C}$
• Force ($F$) = ?

Solution:

According to the definition of electric field intensity:

$$E = \frac{F}{q} \implies F = q \cdot E$$ $$F = (2 \times 10^{-6}\text{ C}) \times (3.42 \times 10^{11}\text{ N/C})$$ $$F = 6.84 \times 10^5\text{ N}$$

RESULT: The electric force acting on the charge is $6.84 \times 10^5\text{ N}$.

Potential & Capacitance

4. What is the charge on the capacitor, if a $40\text{ }\mu\text{F}$ capacitor has a potential difference of $6\text{ V}$ across it?

Data:

• Capacitance ($C$) = $40\text{ }\mu\text{F} = 40 \times 10^{-6}\text{ F}$
• Potential difference ($V$) = $6\text{ V}$
• Charge ($Q$) = ?

Solution:

According to the fundamental equation of capacitance:

$$Q = C \cdot V$$ $$Q = (40 \times 10^{-6}\text{ F}) \times 6\text{ V}$$ $$Q = 240 \times 10^{-6}\text{ C} = 2.4 \times 10^{-4}\text{ C}$$

RESULT: The charge accumulated on the capacitor is $2.4 \times 10^{-4}\text{ C}$ (or $240\text{ }\mu\text{C}$).

5. The potential difference between two points is $100\text{ V}$. If an unknown charge is moved between these points, the amount of work done is $500\text{ J}$. Find the amount of charge.

Data:

• Potential difference ($V$) = $100\text{ V}$
• Work done ($W$) = $500\text{ J}$
• Charge ($Q$) = ?

Solution:

According to the definition of electric potential difference:

$$V = \frac{W}{Q} \implies Q = \frac{W}{V}$$ $$Q = \frac{500\text{ J}}{100\text{ V}} = 5\text{ C}$$

RESULT: The amount of charge moved between the points is $5\text{ C}$.

6. Find the equivalent capacitance when a $4\text{ }\mu\text{F}$, $3\text{ }\mu\text{F}$, and $2\text{ }\mu\text{F}$ capacitor are connected in series.

Data:

• Capacitance ($C_1$) = $4\text{ }\mu\text{F}$
• Capacitance ($C_2$) = $3\text{ }\mu\text{F}$
• Capacitance ($C_3$) = $2\text{ }\mu\text{F}$
• Equivalent Capacitance ($C_{eq}$) = ?

Solution:

When capacitors are connected in series, the reciprocal of equivalent capacitance is equal to the sum of reciprocals of individual capacitances:

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$ $$\frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{3} + \frac{1}{2}$$ $$\text{Taking LCM of 4, 3, and 2, which is 12:}$$ $$\frac{1}{C_{eq}} = \frac{3 + 4 + 6}{12} = \frac{13}{12}$$ $$C_{eq} = \frac{12}{13} \approx 0.923\text{ }\mu\text{F}$$

RESULT: The equivalent capacitance of the series network is $0.923\text{ }\mu\text{F}$.