東京工業大学 工学院 電気電子系 2022年8月実施 電磁気学1 Author Zero
Description 真空中に固定された平行平板コンデンサについて，以下の問に答えよ。真空の誘電率を ε 0 \varepsilon_0 ε 0
(1) 図 1.1 1.1 1.1 a a a b b b x x x
① このコンデンサの静電容量 C 1 C_1 C 1 a , b , x , ε 0 a,b,x,\varepsilon_0 a , b , x , ε 0
② 2 2 2 V V V E 1 E_1 E 1 U 1 U_1 U 1 F 1 F_1 F 1 a , b , x , V , ε 0 a,b,x,V,\varepsilon_0 a , b , x , V , ε 0
③ 図 1.2 1.2 1.2 2 2 2 ± Q \pm Q ± Q E 2 E_2 E 2 U 2 U_2 U 2 F 2 F_2 F 2 a , b , x , Q , ε 0 a,b,x,Q,\varepsilon_0 a , b , x , Q , ε 0
(2) 厚さが無視できる一辺の長さ a a a x x x 1.3 1.3 1.3 a a a d d d ε ( ε > ε 0 ) \varepsilon(\varepsilon > \varepsilon_0) ε ( ε > ε 0 ) l ( 0 < l < a ) l(0 < l < a) l ( 0 < l < a ) V V V
① 図 1.3 1.3 1.3 C 1 C_1 C 1 C 11 C_{11} C 11 a , d , x , l , ε 0 , ε a,d,x,l,\varepsilon_0,\varepsilon a , d , x , l , ε 0 , ε
② 誘電体板に働く静電気力の大きさ
F 3 F_3 F 3 a , d , x , l , V , ε 0 , ε a,d,x,l,V,\varepsilon
_0,\varepsilon a , d , x , l , V , ε 0 , ε 1.3 1.3 1.3
Kai (1)
①
ε 0 E = Q a b ⇒ E = Q ε 0 a b ⇒ V = E x = Q ε 0 a b x \varepsilon_0E = \frac{Q}{ab} \Rightarrow E = \frac{Q}{\varepsilon_0ab} \Rightarrow V = Ex = \frac{Q}{\varepsilon_0ab}x ε 0 E = ab Q ⇒ E = ε 0 ab Q ⇒ V = E x = ε 0 ab Q x 従って、
C 1 = Q V = ε 0 a b x C_1 = \frac{Q}{V} = \frac{\varepsilon_0ab}{x} C 1 = V Q = x ε 0 ab V = Q ε 0 a b x V = \frac{Q}{\varepsilon_0ab}x V = ε 0 ab Q x
Q = ε 0 a b x V Q = \frac{\varepsilon_0ab}{x}V Q = x ε 0 ab V E 1 = 1 ε 0 a b ⋅ ε 0 a b x V = V x E_1 = \frac{1}{\varepsilon_0ab}\cdot \frac{\varepsilon_0ab}{x}V = \frac{V}{x} E 1 = ε 0 ab 1 ⋅ x ε 0 ab V = x V U 1 = 1 2 C 1 V 2 = 1 2 ⋅ ε 0 a b x V 2 U_1 = \frac{1}{2}C_1V^2 = \frac{1}{2} \cdot \frac{\varepsilon_0ab}{x}V^2 U 1 = 2 1 C 1 V 2 = 2 1 ⋅ x ε 0 ab V 2 F 1 = ∂ U 1 ∂ x = − ε 0 a b 2 x 2 V 2 F_1 = \frac{\partial U_1}{\partial x} = -\frac{\varepsilon_0ab}{2x^2}V^2 F 1 = ∂ x ∂ U 1 = − 2 x 2 ε 0 ab V 2 静電気力の向きは導体板が引き合う方向である。
E 2 = Q ε 0 a b E_2 = \frac{Q}{\varepsilon_0ab} E 2 = ε 0 ab Q U 2 = 1 2 ⋅ Q 2 C 1 = 1 2 ⋅ x ε 0 a b Q 2 U_2 = \frac{1}{2} \cdot \frac{Q^2}{C_1} = \frac{1}{2} \cdot \frac{x}{\varepsilon_0ab}Q^2 U 2 = 2 1 ⋅ C 1 Q 2 = 2 1 ⋅ ε 0 ab x Q 2 F 2 = − ∂ U 2 ∂ x = − Q 2 2 ε 0 a b F_2 = -\frac{\partial U_2}{\partial x} = -\frac{Q^2}{2\varepsilon_0ab} F 2 = − ∂ x ∂ U 2 = − 2 ε 0 ab Q 2 静電気力の向きは導体板が引き合う方向である。
(2)
領域 I II にそれぞれ ± Q 1 , ± Q 2 \pm Q_1,\pm Q_2 ± Q 1 , ± Q 2
(i) 領域 I
ε 0 E 1 = Q 1 a ( a − l ) E 1 = Q ε 0 a ( a − l ) V = x E 1 = x ε 0 a ( a − l ) Q 1 \begin{aligned}
\varepsilon_0E_1 &= \frac{Q_1}{a(a - l)} \\
E_1 &= \frac{Q}{\varepsilon_0a(a - l)} \\
V &= xE_1 \\
&= \frac{x}{\varepsilon_0a(a - l)}Q_1
\end{aligned} ε 0 E 1 E 1 V = a ( a − l ) Q 1 = ε 0 a ( a − l ) Q = x E 1 = ε 0 a ( a − l ) x Q 1 従って、
C I = Q 1 V = ε 0 a ( a − l ) x C_I = \frac{Q_1}{V} = \frac{\varepsilon_0a(a - l)}{x} C I = V Q 1 = x ε 0 a ( a − l ) (ii) 領域 II
ε 0 E 21 = Q 2 a l ⇒ E 21 = Q 2 ε 0 a l \varepsilon_0 E_{21} = \frac{Q_2}{al} \Rightarrow E_{21} = \frac{Q_2}{\varepsilon_0al} ε 0 E 21 = a l Q 2 ⇒ E 21 = ε 0 a l Q 2 V 21 = E 21 ( x − d ) = Q 2 ε 0 a l ( x − d ) V_{21} = E_{21}(x - d) = \frac{Q_2}{\varepsilon_0al}(x - d) V 21 = E 21 ( x − d ) = ε 0 a l Q 2 ( x − d ) また、境界条件より、
ε 0 E 21 = ε E 22 \varepsilon_0E_{21} = \varepsilon E_{22} ε 0 E 21 = ε E 22 E 22 = ε 0 ε ⋅ Q 2 ε 0 a l = Q 2 ε a l E_{22} = \frac{\varepsilon_0}{\varepsilon} \cdot \frac{Q_2}{\varepsilon_0al} = \frac{Q_2}{\varepsilon al} E 22 = ε ε 0 ⋅ ε 0 a l Q 2 = ε a l Q 2 V 22 = E 22 d = Q 2 ε a l d V_{22} = E_{22}d = \frac{Q_2}{\varepsilon al}d V 22 = E 22 d = ε a l Q 2 d V = V 21 + V 22 = ( x − d ε 0 a l + d ε a l ) Q 2 = Q 2 a l ( x − d ε 0 + d ε ) V = V_{21} + V_{22} = (\frac{x - d}{\varepsilon_0 al} + \frac{d}{\varepsilon al})Q_2 = \frac{Q_2}{al}(\frac{x-d}{\varepsilon_0} + \frac{d}{\varepsilon}) V = V 21 + V 22 = ( ε 0 a l x − d + ε a l d ) Q 2 = a l Q 2 ( ε 0 x − d + ε d ) 従って、
C I I = Q 2 V = a l x − d ε 0 + d ε C_{II} = \frac{Q_2}{V} = \frac{al}{\frac{x-d}{\varepsilon_0} + \frac{d}{\varepsilon}} C II = V Q 2 = ε 0 x − d + ε d a l C = C I + C I I = ε 0 a ( a − l ) x + a l x − d ε 0 + d ε \begin{aligned}
C &= C_I + C_{II} = \frac{\varepsilon_0a(a - l)}{x} + \frac{al}{\frac{x-d}{\varepsilon_0} + \frac{d}{\varepsilon}}
\end{aligned} C = C I + C II = x ε 0 a ( a − l ) + ε 0 x − d + ε d a l V 3 = 1 2 C V 2 = 1 2 [ ε 0 a ( a − l ) x + a l x − d ε 0 + d ε ] V 2 \begin{aligned}
V_3 &= \frac{1}{2}CV^2 = \frac{1}{2} \bigg[\frac{\varepsilon_0a(a - l)}{x} + \frac{al}{\frac{x-d}{\varepsilon_0} + \frac{d}{\varepsilon}}\bigg]V^2
\end{aligned} V 3 = 2 1 C V 2 = 2 1 [ x ε 0 a ( a − l ) + ε 0 x − d + ε d a l ] V 2 F 3 = ∂ U 3 ∂ l = 1 2 [ − ε 0 a x + a x − d ε 0 + d ε ] V 2 = 1 2 [ − ε 0 a x + ε 0 ε a ε ( x − d ) + ε 0 d ] V 2 = 1 2 [ − ε 0 a x + ε 0 ε a d ( ε 0 − ε ) + ε x ] V 2 = 1 2 ⋅ ε 0 ( ε − ε 0 ) d a x [ d ε 0 + ( x − d ) ε ] V 2 \begin{aligned}
F_3 = \frac{\partial U_3}{\partial l} &= \frac{1}{2} \bigg[-\frac{\varepsilon_0a}{x} + \frac{a}{\frac{x-d}{\varepsilon_0} + \frac{d}{\varepsilon}}\bigg]V^2 \\
&= \frac{1}{2} \bigg[-\frac{\varepsilon_0a}{x} + \frac{\varepsilon_0 \varepsilon a}{\varepsilon(x - d) + \varepsilon_0d}\bigg]V^2 \\
&= \frac{1}{2} \bigg[-\frac{\varepsilon_0a}{x} + \frac{\varepsilon_0 \varepsilon a}{d(\varepsilon_0 - \varepsilon) + \varepsilon x}\bigg]V^2 \\
&= \frac{1}{2} \cdot \frac{\varepsilon_0(\varepsilon - \varepsilon_0)da}{x[d\varepsilon_0 + (x - d)\varepsilon]}V^2
\end{aligned} F 3 = ∂ l ∂ U 3 = 2 1 [ − x ε 0 a + ε 0 x − d + ε d a ] V 2 = 2 1 [ − x ε 0 a + ε ( x − d ) + ε 0 d ε 0 ε a ] V 2 = 2 1 [ − x ε 0 a + d ( ε 0 − ε ) + ε x ε 0 ε a ] V 2 = 2 1 ⋅ x [ d ε 0 + ( x − d ) ε ] ε 0 ( ε − ε 0 ) d a V 2 F 3 > 0 F_3 > 0 F 3 > 0 1.3 1.3 1.3
