Question: EXERCISE 1
1.1 Find the total charge in a particle containing 11 electrons, if the electronic charge is 1.6 x 10^-19 C
1.2 Two-point charges of value 6.3 x 10^-6 C and 7.5 x 10^-6 C are 3.0 cm apart. Find the force between them.
1.3 Determine the electric field at a distance of 2cm on a positive test charge due to a charge of 3.3 x 10^-6 C.
1.4 Three electric charges of magnitude 2C, 3C and 4C are enclosed in a surface. What is the total flux enclosed by the surface.
1.5 Calculate the area of the plates that will produce a 3F capacitor if it is desired to separate the plate by a 6 mm air gap.
1.6 A capacitor is constructed from two metallic plates of dimension 0.3m x 0.4m which are spaced 6 mm apart. Using dry air at its dielectric material, calculate the capacitance of the capacitor. Take permittivity of free space to be 8.85 x 10^-12 F/m^2
1.7 A conducting sphere of diameter 20 cm is charged to a potential of 80,000V with respect to earth potential. Calculate the charge on the sphere assuming it to be situated in air.
1.8 Find the equivalent capacitance of three capacitors of values 0.1bcF, 0.2bcF, and 0.3bcF are connected (i) in series (ii) in parallel
1.9 Three capacitors of values 10bcF, 12bcF and 33bcF are connected together. What is their equivalent capacitance if they are connective in (i) series (ii) parallel
1.10 Calculate the charge stored by a capacitors whose capacitance is 5 bcF when a voltage across the capacitor is 3V
1.11 An electrolytic capacitor has a capacitance of 70 bcF. It is charged through a safety resistor of 80 K9 to a potential difference of 220 V. Calculate (i) the charge on the plate (ii) the electrical energy stored
1.12 Consider a parallel plate capacitor with plate area of 120 cm^2 and plate separation of 1.8 cm. If the potential difference applied between the plates is 52 volts. Find
(a) the capacitance of the capacitor
(b) what free charge appears on the plate
(c) If the battery was disconnected and a dielectric slab of thickness 0.6 mm was inserted between the plates, what is the electric field in the gap between the plates and the dielectric slab if k = 3
(d) what is the electric field in the dielectric slab.
1. **Problem 1.1:** Find the total charge in a particle containing 11 electrons, given the electronic charge is $1.6 \times 10^{-19}$ C.
2. **Formula:** Total charge $Q = n \times e$ where $n$ is the number of electrons and $e$ is the charge of one electron.
3. **Calculation:**
$$Q = 11 \times (-1.6 \times 10^{-19}) = -1.76 \times 10^{-18} \text{ C}$$
The charge is negative because electrons carry negative charge.
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1. **Problem 1.2:** Find the force between two point charges $6.3 \times 10^{-6}$ C and $7.5 \times 10^{-6}$ C separated by $3.0$ cm.
2. **Formula:** Coulomb's law:
$$F = k \frac{|q_1 q_2|}{r^2}$$
where $k = 9 \times 10^9$ Nm$^2$/C$^2$, $r$ in meters.
3. **Calculation:**
Convert $r = 3.0$ cm = $0.03$ m.
$$F = 9 \times 10^9 \times \frac{6.3 \times 10^{-6} \times 7.5 \times 10^{-6}}{(0.03)^2}$$
$$= 9 \times 10^9 \times \frac{4.725 \times 10^{-11}}{9 \times 10^{-4}} = 9 \times 10^9 \times 5.25 \times 10^{-8} = 472.5 \text{ N}$$
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1. **Problem 1.3:** Determine the electric field at $2$ cm from a charge $3.3 \times 10^{-6}$ C.
2. **Formula:** Electric field $E = k \frac{|q|}{r^2}$
3. **Calculation:**
$r = 2$ cm = $0.02$ m.
$$E = 9 \times 10^9 \times \frac{3.3 \times 10^{-6}}{(0.02)^2} = 9 \times 10^9 \times \frac{3.3 \times 10^{-6}}{4 \times 10^{-4}} = 9 \times 10^9 \times 8.25 \times 10^{-3} = 7.425 \times 10^7 \text{ N/C}$$
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1. **Problem 1.4:** Total flux enclosed by a surface with charges $2$ C, $3$ C, and $4$ C.
2. **Formula:** Gauss's law:
$$\Phi = \frac{Q_{enc}}{\epsilon_0}$$
where $Q_{enc} = 2 + 3 + 4 = 9$ C, and $\epsilon_0 = 8.85 \times 10^{-12}$ C$^2$/N m$^2$.
3. **Calculation:**
$$\Phi = \frac{9}{8.85 \times 10^{-12}} = 1.017 \times 10^{12} \text{ Nm}^2/\text{C}$$
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1. **Problem 1.5:** Calculate area $A$ for a capacitor of capacitance $3$ F with plate separation $d = 6$ mm.
2. **Formula:**
$$C = \epsilon_0 \frac{A}{d} \Rightarrow A = \frac{C d}{\epsilon_0}$$
3. **Calculation:**
$d = 6$ mm = $6 \times 10^{-3}$ m.
$$A = \frac{3 \times 6 \times 10^{-3}}{8.85 \times 10^{-12}} = 2.03 \times 10^{9} \text{ m}^2$$
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1. **Problem 1.6:** Capacitance of plates $0.3$ m x $0.4$ m spaced $6$ mm apart, air dielectric.
2. **Formula:**
$$C = \epsilon_0 \frac{A}{d}$$
3. **Calculation:**
$$A = 0.3 \times 0.4 = 0.12 \text{ m}^2$$
$$d = 6 \times 10^{-3} \text{ m}$$
$$C = 8.85 \times 10^{-12} \times \frac{0.12}{6 \times 10^{-3}} = 1.77 \times 10^{-10} \text{ F}$$
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1. **Problem 1.7:** Charge on a conducting sphere diameter $20$ cm, potential $80,000$ V.
2. **Formula:**
$$V = \frac{k Q}{r} \Rightarrow Q = \frac{V r}{k}$$
3. **Calculation:**
$$r = 0.1 \text{ m}, k = 9 \times 10^9$$
$$Q = \frac{80,000 \times 0.1}{9 \times 10^9} = 8.89 \times 10^{-7} \text{ C}$$
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1. **Problem 1.8:** Equivalent capacitance of $0.1 \mu F$, $0.2 \mu F$, $0.3 \mu F$ capacitors.
2. **Series:**
$$\frac{1}{C_s} = \frac{1}{0.1} + \frac{1}{0.2} + \frac{1}{0.3} = 10 + 5 + 3.33 = 18.33$$
$$C_s = \frac{1}{18.33} = 0.0545 \mu F$$
3. **Parallel:**
$$C_p = 0.1 + 0.2 + 0.3 = 0.6 \mu F$$
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1. **Problem 1.9:** Equivalent capacitance of $10 \mu F$, $12 \mu F$, $33 \mu F$ capacitors.
2. **Series:**
$$\frac{1}{C_s} = \frac{1}{10} + \frac{1}{12} + \frac{1}{33} = 0.1 + 0.0833 + 0.0303 = 0.2136$$
$$C_s = \frac{1}{0.2136} = 4.68 \mu F$$
3. **Parallel:**
$$C_p = 10 + 12 + 33 = 55 \mu F$$
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1. **Problem 1.10:** Charge stored by capacitor $5 \mu F$ at $3$ V.
2. **Formula:**
$$Q = C V$$
3. **Calculation:**
$$Q = 5 \times 10^{-6} \times 3 = 1.5 \times 10^{-5} \text{ C}$$
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1. **Problem 1.11:** Electrolytic capacitor $70 \mu F$, resistor $80$ k$\Omega$, voltage $220$ V.
2. **Charge:**
$$Q = C V = 70 \times 10^{-6} \times 220 = 0.0154 \text{ C}$$
3. **Energy stored:**
$$E = \frac{1}{2} C V^2 = 0.5 \times 70 \times 10^{-6} \times 220^2 = 1.694 \text{ J}$$
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1. **Problem 1.12:** Parallel plate capacitor with area $120$ cm$^2$, separation $1.8$ cm, voltage $52$ V.
2. **(a) Capacitance:**
$$A = 120 \text{ cm}^2 = 120 \times 10^{-4} = 0.012 \text{ m}^2$$
$$d = 1.8 \text{ cm} = 0.018 \text{ m}$$
$$C = \epsilon_0 \frac{A}{d} = 8.85 \times 10^{-12} \times \frac{0.012}{0.018} = 5.9 \times 10^{-12} \text{ F}$$
3. **(b) Free charge:**
$$Q = C V = 5.9 \times 10^{-12} \times 52 = 3.07 \times 10^{-10} \text{ C}$$
4. **(c) Electric field in gap and dielectric slab:**
Battery disconnected means charge $Q$ constant.
Thickness dielectric $t = 0.6$ mm = $0.0006$ m, gap thickness $d_g = 0.018 - 0.0006 = 0.0174$ m.
Dielectric constant $k=3$.
Electric field in dielectric slab:
$$E_d = \frac{Q}{\epsilon_0 k A}$$
Electric field in gap:
$$E_g = \frac{Q}{\epsilon_0 A}$$
Calculate $E_g$:
$$E_g = \frac{Q}{\epsilon_0 A} = \frac{3.07 \times 10^{-10}}{8.85 \times 10^{-12} \times 0.012} = 2.89 \times 10^{3} \text{ V/m}$$
Calculate $E_d$:
$$E_d = \frac{E_g}{k} = \frac{2.89 \times 10^{3}}{3} = 963 \text{ V/m}$$
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**Final answers:**
1.1 $Q = -1.76 \times 10^{-18}$ C
1.2 $F = 472.5$ N
1.3 $E = 7.425 \times 10^{7}$ N/C
1.4 $\Phi = 1.017 \times 10^{12}$ Nm$^2$/C
1.5 $A = 2.03 \times 10^{9}$ m$^2$
1.6 $C = 1.77 \times 10^{-10}$ F
1.7 $Q = 8.89 \times 10^{-7}$ C
1.8 (i) $0.0545 \mu F$, (ii) $0.6 \mu F$
1.9 (i) $4.68 \mu F$, (ii) $55 \mu F$
1.10 $Q = 1.5 \times 10^{-5}$ C
1.11 (i) $0.0154$ C, (ii) $1.694$ J
1.12 (a) $5.9 \times 10^{-12}$ F, (b) $3.07 \times 10^{-10}$ C, (c) $E_g = 2.89 \times 10^{3}$ V/m, (d) $E_d = 963$ V/m