Question

2025 - 2026 - 1 exercise for chapter 24 student id no. s(12 - 10060) name abigail awusu 1. fig. 1 shows 3 capacitors, of equal capacitance c, connected to a battery of voltage v. what is the equivalent capacitance of this combination? ( ) (a) 3c/2 (b) c/2 (c) 2c/3 (d) 3c 2. what charge appears on the plates of a 6.0 - μf capacitor when it is charged to 40 v? ( ) (a) 60 μc (b) 120 μc (c) 180 μc (d) 240 μc 3. the magnitude of the charge on each plate of a parallel plate capacitor is 4 μc and the potential difference between the plates is 80 v. what is the capacitance of this capacitor? ( ) (a) 300×10^(-6) f (b) 100×10^(-6) f (c) 20×10^(-6) f (d) 5×10^(-8) f 4. a parallel - plate capacitor has plates of area 0.80 m² separated by a distance of 3.0 mm. what is this capacitors capacitance? ( ) (a) 270 f (b) 70 f (c) 2.4×10^(-9) f (d) 4.8×10^(-10) f 5. a parallel - plate capacitor has a voltage v = 6.0 v between its plates. each plate carries a surface charge density σ = 7.0 nc/m². what is the separation of the plates? ( ) (a) 2.5 mm (b) 5.6 mm (c) 7.6 mm (d) 8.2 mm

Explanation:

Step1: Recall capacitor - combination formula

For capacitors in parallel, the equivalent capacitance $C_{eq}=\sum_{i = 1}^{n}C_i$. For capacitors in series, $\frac{1}{C_{eq}}=\sum_{i = 1}^{n}\frac{1}{C_i}$. In Fig. 1, two capacitors are in parallel and then in series with the third one. The two - parallel capacitors have an equivalent capacitance of $C_{p}=C + C=2C$. Then, when this $2C$ capacitor is in series with the third $C$ capacitor, $\frac{1}{C_{eq}}=\frac{1}{2C}+\frac{1}{C}=\frac{1 + 2}{2C}=\frac{3}{2C}$, so $C_{eq}=\frac{2C}{3}$.

Step2: Recall the formula $Q = CV$

For question 2, given $C = 6.0\ \mu F$ and $V = 40\ V$, using the formula $Q=CV$, we have $Q=(6.0\times10^{- 6}\ F)\times40\ V = 240\times10^{-6}\ C=240\ \mu C$.

Step3: Recall the formula $C=\frac{Q}{V}$

For question 3, given $Q = 4\ \mu C=4\times10^{-6}\ C$ and $V = 80\ V$, using the formula $C=\frac{Q}{V}$, we get $C=\frac{4\times10^{-6}\ C}{80\ V}=5\times10^{-8}\ F$.

Step4: Recall the formula for parallel - plate capacitor $C=\frac{\epsilon_0A}{d}$

For question 4, with $\epsilon_0 = 8.85\times10^{-12}\ F/m$, $A = 0.80\ m^{2}$ and $d = 3.0\times10^{-3}\ m$, $C=\frac{8.85\times10^{-12}\ F/m\times0.80\ m^{2}}{3.0\times10^{-3}\ m}=2.4\times10^{-9}\ F$.

Step5: Recall the relationships $E=\frac{\sigma}{\epsilon_0}$ and $V = Ed$

First, $E=\frac{\sigma}{\epsilon_0}$, where $\sigma = 7.0\times10^{-9}\ C/m^{2}$ and $\epsilon_0 = 8.85\times10^{-12}\ F/m$, so $E=\frac{7.0\times10^{-9}\ C/m^{2}}{8.85\times10^{-12}\ F/m}$. Then, since $V = Ed$ and $V = 6.0\ V$, $d=\frac{V\epsilon_0}{\sigma}=\frac{6.0\ V\times8.85\times10^{-12}\ F/m}{7.0\times10^{-9}\ C/m^{2}}\approx7.6\times10^{-3}\ m = 7.6\ mm$.

Answer:

1. C. $2C/3$
2. D. $240\ \mu C$
3. D. $5\times 10^{-8}\ F$
4. C. $2.4\times 10^{-9}\ F$
5. C. $7.6\ mm$