Video: Determining the Force on a Charge due to a Second Charge

Suppose the parallel plate capacitor shown is accumulating charge at a rate of 0.035 C/s. What is the magnitude of the resulting induced magnetic field at a distance of 15 cm from the capacitor?

02:06

Video Transcript

Suppose the parallel plate capacitor shown is accumulating charge at a rate of 0.035 coulombs per second. What is the magnitude of the resulting induced magnetic field at a distance of 15 centimeters from the capacitor?

We can label this magnitude of magnetic field capital ๐ต. And as we study our diagram, we see that a current, ๐ผ, is flowing into our parallel plate capacitor and then out the other side. And we wanna solve for the magnetic field thatโ€™s created a distance of 15 centimeters away from the center of these places.

Since the distance at which we wanna solve for the magnetic field is outside of our plates, that is, greater than our platesโ€™ radius, then the magnetic field at this point is given by the relationship ๐œ‡ naught times ๐ผ over two ๐œ‹๐‘Ÿ, where ๐‘Ÿ is the distance from the center of the plates. The value ๐œ‡ naught is a constant called the permeability of free space. This constant has a value of 1.26 times 10 to the negative sixth tesla meters per ampere.

Looking then at our equation for the magnetic field magnitude ๐ต, we know ๐œ‡ naught, and ๐‘Ÿ is given to us as 15 centimeters. That means we just need to solve for the current, ๐ผ, and then weโ€™ll be able to solve for the magnetic field.

We can recall that an ampere, which is the unit of current, is equal to a coulomb of charge passing a given point every second. And in our problem statement, weโ€™re told this rate of charge per second that enters the plates. Since a coulomb per second is also an ampere, that means weโ€™ve been told the current.

We can now plug in our values, 1.26 times 10 to the negative sixth tesla meters per amp for ๐œ‡ naught, 0.035 amperes for our current ๐ผ, and, converting to units of meters, 0.15 meters for ๐‘Ÿ. To two significant figures, this equals 4.7 times 10 to the negative eighth tesla. Thatโ€™s the magnetic field strength produced by this parallel plate capacitor the given distance away from the plates.

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