Charge is flowing onto one plate of a parallel-plate capacitor and off the other plate at a rate of 2.50 A.The rate at which the electric field between the plates is changing is approximately A)5.65 $\times$ 10¹¹ N · m²/(C · s) B)11.3 $\times$ 10¹¹ N · m²/(C · s) C)2.82 $\times$ 10¹¹ N · m²/(C · s) D)12.4 $\times$ 10¹¹ N · m²/(C · s) E)1.76 $\times$ 10¹¹ N · m²/(C · s)

Question

Quizplus · Accepted Answer

The rate at which the electric field between the plates is changing is given by:$$\frac{dE}{dt}=\frac{d}{dt}\left(\frac{Q}{\epsilon_0A}ight)=\frac{1}{\epsilon_0A}\frac{dQ}{dt}$$where:* $E$ is the electric field between the plates* $Q$ is the charge on one of the plates* $\epsilon_0$ is the permittivity of free space* $A$ is the area of each plateSubstituting the given values into the equation, we get:$$\frac{dE}{dt}=\frac{1}{8.85	imes10^{-12}	ext{ F/m}^2(0.025	ext{ m})^2}(2.50	ext{ A})$$$$\frac{dE}{dt}=1.13	imes10^{11}	ext{ N}\cdot	ext{m}^2/	ext{C}\cdot	ext{s}$$Therefore, the correct answer is A.

Charge Is Flowing onto One Plate of a Parallel-Plate Capacitor ×\times×

Charge Is Flowing onto One Plate of a Parallel-Plate Capacitor $\times$