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The induced magnetic field at a radial distance 6.0 mm from the central axis of a circular parallel-plate capacitor is 2.0×10-7T. The plates have a radius 3.0 mm. At what ratedE⇶Ä/dtis the electric field between the plates changing?

Short Answer

Expert verified

The value ofdE⇶Ädtbetween the charging plates is2.4×1013V/m·s.

Step by step solution

01

Identification of the given data

The given data is listed as follows,

  • The radial distance of induced magnetic field from the central axis of a circular parallel plate capacitor is,r=6.0mm×1m1000mm=6.0×10-3m
  • The induced magnetic field is,B=2.0×10-7T
  • The radius of the plate is,R=3.0mm×1m1000mm=3.0×10-3m
02

Expression for the Maxwell’s law of induction

The expression for Maxwell’s law of induction is as follows,

∫B⇶Ä·dA⇶Ä=μ0ε0AdEdt

Here, B is the induced magnetic field, A is the area enclosed which is given by the following expression A=πR2(Here, R is the radius of the plate),μ0is the magnetic permeability of the free space with the value of 4π×10-7N/A2,ε0, is the electric permeability of the free space with the value of 8.85×10-12C2/N·m2, anddEdtis then the rate of change of electric field.

03

Determination of the value of dE⇀/dt between the charging plates

Substitute the2Ï€°ùfor dA, andÏ€¸é2 for A in the Maxwell’s law of induction and rearrange the expression.

B2Ï€°ù=μ0E0Ï€R2dEdtB=μ0E0Ï€R22rdEdtdEdt=2Brμ0E0R2

Substitute all the values in the above expression.

dEdt=2×2.0×10-7T×6.0×10-3m4π×10-7N/A28.85×10-12C2/N·m23.0×10-3m2=24.0×10-101001×10-25T·m·A2/C2×1N/A·m1T=2.3975×1013N·A/C2×1C/s·m1A=2.4×1013N/C·s×1V/m1N/C=2.4×1013V/m·s

Thus, the value ofdE⇶Ädtbetween the charging plates is 2.4×1013V/m·s.

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