Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 20, Problem 85P

A capacitor is constructed from two square, metallic plates of sides and separation d. Charges +Q and −Q are placed on the plates, and the power supply is then removed. A material of dielectric constant K is inserted a distance x into the capacitor as shown in Figure P20.85. Assume d is much smaller than x. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor. (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. (d) Obtain a numerical value for the force when x = /2, assuming = 5.00 cm, d = 2.00 mm, the dielectric is glass (κ = 4.50), and the capacitor was charged to 2.00 × 103 V before the dielectric was inserted. Suggestion: The system can be considered as two capacitors connected in parallel.

Figure P20.85

Chapter 20, Problem 85P, A capacitor is constructed from two square, metallic plates of sides  and separation d. Charges +Q

(a)

Expert Solution
Check Mark
To determine

The equivalent capacitance of the device.

Answer to Problem 85P

The equivalent capacitance of the device is ε0ld[l+x(κ1)].

Explanation of Solution

Write the expression for the equivalent capacitance.

Ceq=C1+C2        (I)

Here, Ceq is the equivalent capacitance of the circuit and C1,C2 are capacitance connected in the circuit.

Write the expression for the area with dielectrics.

A=lx        (II)

Here, A is the area, l is the length and x is the width.

Write the equation for capacitance by using equation (II).

  C1=κε0lxd        (III)

Here, k is the dielectric, ε0 is the permittivity of free space, A is the area and d is distance.

Write the expression for the area without dielectrics.

A=l(lx)        (IV)

Write the equation for capacitance by using equation (IV).

  C2=ε0l(lx)d        (V)

Conclusion:

Substitute, κε0lxd for C1, ε0l(lx)d for C2 in Equation (I) to find Ceq.

  Ceq=[κε0lxd]+[ε0l(lx)d]=ε0ld[l+x(κ1)]

Thus, the equivalent capacitance of the device is ε0ld[l+x(κ1)].

(b)

Expert Solution
Check Mark
To determine

The energy stored in the capacitor.

Answer to Problem 85P

The energy stored in the capacitor is Q2d2ε0l[l+x(κ1)]_.

Explanation of Solution

Write the expression for the stored energy.

U=Q22C        (VI)

Here, Q is the charge and U is the energy.

Conclusion:

Substitute, ε0ld[l+x(κ1)] for C in Equation (VI) to find U.

U=Q22ε0ld[l+x(κ1)]=Q2d2ε0l[l+x(κ1)]

Thus, the energy stored in the capacitor is Q2d2ε0l[l+x(κ1)]_.

(c)

Expert Solution
Check Mark
To determine

The direction and the magnitude of the force exerted by the plates on the dielectrics.

Answer to Problem 85P

The direction and the magnitude of the force exerted by the plates on the dielectrics is Q2d(κ1)2ε0l[l+x(κ1)2]_ and directed along positive x-axis.

Explanation of Solution

Write the expression exerted force.

F=dUdx        (V)

Conclusion:

Substitute, Q2d2ε0l[l+x(κ1)] for U in Equation (V) to find F.

  F=d[Q2d2ε0l[l+x(κ1)]]dx=Q2d(κ1)2ε0l[l+x(κ1)2]        (VI)

For x=0, the value of force is,

F=Q2d(κ1)2ε0l[l+(0)(κ1)2]=Q2d(κ1)2ε0l3

For x=l, the value of force is,

F=Q2d(κ1)2ε0l[l+l(κ1)2]Q2d(κ1)2ε0l3κ2

The force is directed along positive x-direction.

Thus, the direction and the magnitude of the force exerted by the plates on the dielectrics is Q2d(κ1)2ε0l[l+x(κ1)2]_ and directed along positive x-axis.

(d)

Expert Solution
Check Mark
To determine

The numerical value of force.

Answer to Problem 85P

The numerical value of force is 205μN_.

Explanation of Solution

Write the expression for the area.

A=l2        (VII)

Write the expression for the initial capacitance.

C0=ε0l2d        (VIII)

Here, C0 is the initial capacitance.

Write the expression for the charge.

Q=C0(ΔV)=ε0l2(ΔV)d        (IX)

Write the expression for the given condition.

x=l2        (X)

Write the expression for the force by using (VI), (VII), (VIII), (IX) and (X).

F=2ε0l(ΔV)2(κ1)d(κ+1)2        (XI)

Conclusion:

Substitute, 8.85×1012C2/N-m2 for ε0, 5.00cm for l, 2.00mm for d, 4.50 for κ, 2.00×103V for (ΔV) in Equation (XI) to find F.

  F=2(8.85×1012C2/N-m2)[(5.00cm)(1×102m1cm)](2.00×103V)2(4.501)[(2.00mm)(1×103m1mm)](4.50+1)2=205×106N=205μN

Thus, the numerical value of force is 205μN_.

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Chapter 20 Solutions

Principles of Physics: A Calculus-Based Text

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