Question

1. a 4 - μf capacitor has a potential drop of 20 v between its plates. the electric potential energy stored in this capacitor is ( )

(a) 0.8 μj (b) 8 μj (c) 80 μj (d) 800 μj

1. a parallel - plate capacitor has a potential difference between the plates of 80 v. if the charge on one of the plates of the capacitor is + 8.0 μc, what is the electrical energy stored by this capacitor? ( )

(a) 640×10⁻⁶ j (b) 320×10⁻⁶ j (c) 5.0×10⁻⁴ j (d) 6.0×10⁻⁴ j

1. two large non - conducting plates of surface area a = 0.25 m² carry equal but opposite charges q = 75 μc. what is the energy density of the electric field between the two plates? ( )

(a) 1.7×10⁷ j/m³ (b) 5.1×10¹³ j/m³ (c) 3.4×10⁷ j/m³ (d) 2.5×10¹¹ j/m³

1. a parallel - plate capacitor stores a charge q = 4.00 nc when connected to a battery of voltage v = 10.0 v. the energy density is then u = 3.62×10⁻⁴ j/m³. what is the surface area of the plates? ( )

(a) 0.0250 m² (b) 0.0500 m² (c) 0.0923 m² (d) 0.1030 m²

Explanation:

Step1: Recall energy - capacitor formula

The electric - potential energy stored in a capacitor is given by $U=\frac{1}{2}CV^{2}=\frac{1}{2}QV=\frac{Q^{2}}{2C}$.

Step2: Solve problem 10

Given $C = 4\ \mu F=4\times10^{- 6}F$ and $V = 20V$. Using $U=\frac{1}{2}CV^{2}$, we have $U=\frac{1}{2}\times(4\times10^{-6}F)\times(20V)^{2}=\frac{1}{2}\times4\times10^{-6}\times400J = 8\times10^{-4}J=800\ \mu J$. So the answer to problem 10 is D.

Step3: Solve problem 11

Given $V = 80V$ and $Q = 8.0\ \mu C=8\times10^{-6}C$. Using $U=\frac{1}{2}QV$, we have $U=\frac{1}{2}\times(8\times10^{-6}C)\times80V=320\times10^{-6}J = 320\ \mu J=3.2\times10^{-4}J$. So the answer to problem 11 is B.

Step4: Recall energy - density formula for parallel - plate capacitor

The electric - field energy density $u=\frac{\epsilon_{0}E^{2}}{2}$, and for a parallel - plate capacitor $E=\frac{\sigma}{\epsilon_{0}}$, where $\sigma=\frac{Q}{A}$. First, find $\sigma=\frac{Q}{A}=\frac{75\times10^{-6}C}{0.25m^{2}} = 3\times10^{-4}C/m^{2}$. Then $E=\frac{\sigma}{\epsilon_{0}}$, and $u=\frac{\epsilon_{0}E^{2}}{2}=\frac{\sigma^{2}}{2\epsilon_{0}}$. Substituting $\epsilon_{0}=8.85\times10^{-12}C^{2}/(N\cdot m^{2})$ and $\sigma = 3\times10^{-4}C/m^{2}$, we get $u=\frac{(3\times10^{-4}C/m^{2})^{2}}{2\times8.85\times10^{-12}C^{2}/(N\cdot m^{2})}\approx5.1\times10^{13}J/m^{3}$. So the answer to problem 12 is B.

Step5: Recall relevant formulas for parallel - plate capacitor

First, from $Q = CV$, we can find $C=\frac{Q}{V}=\frac{4.00\times10^{-9}C}{10.0V}=4\times10^{-10}F$. The energy density $u=\frac{\epsilon_{0}E^{2}}{2}$, and for a parallel - plate capacitor $C=\frac{\epsilon_{0}A}{d}$, $E=\frac{V}{d}$. Also, $u=\frac{\epsilon_{0}E^{2}}{2}=\frac{\epsilon_{0}(V/d)^{2}}{2}$. And $C=\frac{\epsilon_{0}A}{d}$, so $d=\frac{\epsilon_{0}A}{C}$. Substituting $d$ into the energy - density formula: $u=\frac{C^{2}V^{2}}{2\epsilon_{0}A^{2}}$. Rearranging for $A$, we have $A=\sqrt{\frac{C^{2}V^{2}}{2\epsilon_{0}u}}$. Substituting $C = 4\times10^{-10}F$, $V = 10V$, $\epsilon_{0}=8.85\times10^{-12}C^{2}/(N\cdot m^{2})$, and $u = 3.62\times10^{-4}J/m^{3}$, we get $A = 0.0500m^{2}$. So the answer to problem 13 is B.

Answer:

1. D. $800\ \mu J$
2. B. $320\times10^{-4}J$
3. B. $5.1\times10^{13}J/m^{3}$
4. B. $0.0500m^{2}$