Question
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Chapter 33, Problem 1P

(a)

To determine

The time rate of increase of electric field between the plates.

(a)

Expert Solution
Check Mark

Answer to Problem 1P

The time rate of increase of electric field between the plates is 7.19×1011V/ms .

Explanation of Solution

Given info: The current is 0.200A , the radius of the circular plate is 10.0cm and the separation distance between the plates is 4.00mm .

The value of permittivity of free space is 8.85×1012F/m and the value of the permeability constant is ×107H/m

The area of the circular plate is,

A=πr2

Here,

r is the radius of the circular plate.

The formula for the capacitance of the capacitor is.

C=Aε0d=πr2ε0d

Here,

ε0 is the permittivity of free space.

d is the separation between the plates.

Substitute 10.0cm for r , 8.85×1012F/m for ε0 and 4.00mm for d in above equation to find C .

C=π(10.0cm(102m1cm))2(8.85×1012F/m)4.00mm(103m1mm)=6.95×1011F

Thus, the capacitance of the capacitor is 6.95×1011F .

The formula for the capacitance of the capacitor is.

C=QVV=QC

Here,

Q is the charge of the plates.

V is the potential of the plates.

Differentiate the above equation with respect to time.

dVdt=dQdtC

The formula for the current is,

I=dQdt

Then,

dVdt=IC

The formula for the electric field is,

E=Vd

Differentiate the above equation with respect to time.

dEdt=1ddVdt

Substitute IC for dVdt in above equation.

dEdt=ICd

Substitute 0.200A for I , 6.95×1011F for C and 4.00mm for d in above equation to find dEdt .

dEdt=0.200A(6.95×1011F)(4.00mm(103m1mm))=7.19×1011V/ms

Conclusion:

Therefore, the time rate of increase of electric field between the plates is 7.19×1011V/ms .

(b)

To determine

The magnetic field between the plates 5.00cm from the centre.

(b)

Expert Solution
Check Mark

Answer to Problem 1P

The magnetic field between the plates 5.00cm from the centre is 2.0×107T .

Explanation of Solution

Given info: The current is 0.200A , the radius of the circular plate is 10.0cm and the separation distance between the plates is 4.00mm .

The formula for the electric flux through the circular plate is,

ΦE=EAdΦEdt=AdEdt=πr2dEdt

The Maxwell’s equation is.

Bds=ε0μ0dΦEdtB(2πr)=ε0μ0πr2dEdtB=(rε0μ02)dEdt

Substitute 5.00cm for r , 8.85×1012F/m for ε0 , ×107H/m for μ0 and 7.19×1011V/ms for dEdt in above equation to find B .

B=((5.00cm(102m1cm))(8.85×1012F/m)(×107H/m)2)(7.19×1011V/ms)=1.9990×107T2.0×107T

Conclusion:

Therefore, the magnetic field between the plates 5.00cm from the centre is 2.0×107T .

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Chapter 33 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Multi-term

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