Show that the quantity ϵ 0 (dΦ e /dt) has units of current.

Hello, let's go through this practice problem. The Amp Maxwell law can be expressed as follows. The closed line integral of B dot Dr is equal to the magnetic constant, New knot multiplied by the enclosed current plus the electric constant, epsilon knot multiplied by the derivative with respect to time of the double integral of E dot D A. What is the unit of the quantity epsilon knot multiplied by the derivative with respect to time of the double integral of E dot D A option AC coons, option BV volts, option C A and option D volts per second or V over S. This is kind of an interesting problem because even if you don't fully understand the Mp Maxwell Law, the crux of this problem really just comes down to an understanding of how units work and how they multiply together or how they change when we're looking at integrals and derivatives. First off recall that epsilon knot, the electric constant has units of Coolum squared per Newton meters squared. So those are the unit of our first term and this is being multiplied by whatever the units are of the derivative with respect to time of that double integral. So let's write this down as a mathematical expression. So Coolum squared and, and we're, we're gonna ignore value is we're not going to write out the value itself of the electric constant because that's not really relevant when we're just analyzing the units. So the Coolum squared per Newton meter squared and this is being multiplied by a derivative with respect to time. Now, the way units work when we're dealing with derivatives is that the unit of a derivative is just equal to the units of whatever is being differentiated but then divided by the variable of differentiation. So in our case, we have a derivative with respect to time. So the way we handle the units to this is that we take the units of whatever is inside the derivative and then divide it by the units for time which is seconds. So we're taking the units for the electric constant multiplying it by the units of whatever is inside the derivative. But then dividing it by the unit of time which is seconds. Now the contents of that derivative is an integral. And the way units work for an integral is fairly straightforward, we just take the units together of whatever is being integrated. And don't forget to ignore the variable of integration. So the units for this integral which is E dot D A is going to be the units of the electric field multiplied by the units of area. So recall that electric field has units of Newton's per Coolum and units of area is meters squared. So these are the units of our expression. And now we can do some simplification in canceling things, some things out. We can see that there is a new ones that will cancel out that is in the numerator and the denominator, the meters squared will also cancel out since there's one in the numerator and the denominator and one of the Coombs will cancel out because we have a Coolum squared in the numerator. And then another one that is being divided. So what we're left with is units of coulombs per second or charge per unit time, which is the definition of current and current has units of pi. So this is the equivalent of an pi so the units for the quantity is, which corresponds to option C and that is it for this problem. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

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31. Alternating Current

Capacitors in AC Circuits

Physics

31. Alternating Current

Capacitors in AC Circuits

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Problem 9

Textbook Question

Show that the quantity ϵ₀(dΦ_e/dt) has units of current.

Verified step by step guidance

Step 1: Begin by identifying the physical quantities involved. Here, ϵ₀ represents the permittivity of free space, dΦₑ/dt is the rate of change of electric flux, and we need to show that the product ϵ₀(dΦₑ/dt) has units of current.

Step 2: Recall the SI unit of permittivity of free space (ϵ₀). It is measured in farads per meter (F/m). A farad (F) is equivalent to coulombs per volt (C/V), so ϵ₀ has units of C/(V·m).

Step 3: Consider the electric flux Φₑ. Electric flux is defined as the integral of the electric field over an area, Φₑ = ∫E·dA. The electric field (E) has units of volts per meter (V/m), and area (A) has units of square meters (m²). Therefore, Φₑ has units of V·m.

Step 4: The rate of change of electric flux, dΦₑ/dt, involves dividing the units of Φₑ (V·m) by time (seconds, s). Thus, dΦₑ/dt has units of V·m/s.

Step 5: Multiply ϵ₀ (C/(V·m)) by dΦₑ/dt (V·m/s). The units become (C/(V·m)) × (V·m/s) = C/s. Since coulombs per second (C/s) is the unit of electric current, this demonstrates that ϵ₀(dΦₑ/dt) has units of current.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Flux (Φe)

Electric flux (Φe) is a measure of the electric field passing through a given area. It is calculated as the product of the electric field (E) and the area (A) through which it passes, taking into account the angle between the field lines and the normal to the surface. The unit of electric flux is the volt-meter (V·m), which can also be expressed in terms of base SI units as kg·m²/(s³·A).

Permittivity of Free Space (ϵ0)

The permittivity of free space (ϵ0) is a fundamental physical constant that characterizes how electric fields interact with the vacuum of space. It has a value of approximately 8.85 x 10^-12 F/m (farads per meter) and is crucial in determining the capacitance of capacitors and the behavior of electric fields in free space. This constant plays a key role in Maxwell's equations, which describe electromagnetism.

Current (I)

Electric current (I) is the flow of electric charge through a conductor, typically measured in amperes (A). One ampere is defined as one coulomb of charge passing through a point in a circuit per second. In the context of the equation, showing that ϵ0(dΦe/dt) has units of current involves demonstrating that the units of the expression correspond to those of amperes, linking electric flux change over time to charge flow.

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