Question

4 $C = \frac{\epsilon_0 A}{d}$ $\epsilon_0 = 8.85 \times 10^{-12}$ $L = 12 cm \rightarrow 0.12 m$ a) $d = 0.5 mm \rightarrow 0.0005 m$ $A = L^2 = (0.12 m)^2 = 0.0144 m^2$ $C = \frac{(8.85 \times 10^{-12})(0.0144)}{0.0005}$ $C = 2.55 \times 10^{-10} F$ $Q = CV$ b) $Q = 2.55 \times 10^{-10} \times 200 V = Q = 5.1 \times 10^{-8} C$ c) $U = \frac{Q^2}{2C} = \frac{1}{2} CV^2 \rightarrow U = \frac{1}{2} (2.55 \times 10^{-10})(200)^2$ $\frac{1}{2} (2.55 \times 10^{-10})(40000) = \approx 5.1 \times 10^{-6} J$

Name: 4 c fracepsilon0 ad epsilon0 885 times 10 12 l 12 cm rightarrow 012 m a d 05 mm rightarrow 00005 m a l2 012 m2 00144 m2 c frac885 times 10 120014400005 c 255 times 10 10 f q cv b q 255 tim 28736
Uploaded: 2023-06-24T10:28:39-08:00
Duration: 2 min 39 s
Channel: Adi S
Description: 4 c fracepsilon0 ad epsilon0 885 times 10 12 l 12 cm rightarrow 012 m a d 05 mm rightarrow 00005 m a l2 012 m2 00144 m2 c frac885 times 10 120014400005 c 255 times 10 10 f q cv b q 255 tim 28736

          4 $C = \frac{\epsilon_0 A}{d}$
$\epsilon_0 = 8.85 \times 10^{-12}$
$L = 12 cm \rightarrow 0.12 m$
a) $d = 0.5 mm \rightarrow 0.0005 m$
$A = L^2 = (0.12 m)^2 = 0.0144 m^2$
$C = \frac{(8.85 \times 10^{-12})(0.0144)}{0.0005}$
$C = 2.55 \times 10^{-10} F$
$Q = CV$
b) $Q = 2.55 \times 10^{-10} \times 200 V = Q = 5.1 \times 10^{-8} C$
c) $U = \frac{Q^2}{2C} = \frac{1}{2} CV^2 \rightarrow U = \frac{1}{2} (2.55 \times 10^{-10})(200)^2$
$\frac{1}{2} (2.55 \times 10^{-10})(40000) = \approx 5.1 \times 10^{-6} J$

$4 c fracepsilon0 ad epsilon0 885 times 10 12 l 12 cm rightarrow 012 m a d 05 mm rightarrow 00005 m a l2 012 m2 00144 m2 c frac885 times 10 120014400005 c 255 times 10 10 f q cv b q 255 tim 28736$

Added by Sebastian F.

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