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 82P

(a)

To determine

The expression for the electric potential.

(a)

Expert Solution
Check Mark

Answer to Problem 82P

The expression for the electric potential is V=kpcosθr2.

Explanation of Solution

The diagram for the system is given by figure 1.

Principles of Physics: A Calculus-Based Text, Chapter 20, Problem 82P

Write the expression total potential.

V=V1V2        (I)

Here, V is the total charge and V1,V2 are the potential differences at two points.

Write the equation for potential difference at first point.

  V1=kqr1        (II)

Here, k is the Coulomb’s constant, q is the charge and r1 is the distance.

Write the equation for potential difference at second point.

  V2=kqr2        (II)

Here, k is the Coulomb’s constant, q is the charge and r2 is the distance.

For r>>a write the expression for the difference between the distances.

(r2r1)2acosθ        (III)

Here. a is the distance.

The distance between the points are almost same, r1r2.

Write the expression for dipole.

p=2aq        (IV)

Conclusion:

Substitute, kqr1 for V1, kqr2 for V2, 2acosθ for (r2r1), r for r1, r for r2, p for 2aq in Equation (I) to find V.

  V=(kqr1)(kqr2)=kqr1r2(r2r1)=k(pcosθ)r2

Thus, the expression for the electric potential is V=kpcosθr2.

(b)

To determine

The radial and perpendicular component of electric field.

(b)

Expert Solution
Check Mark

Answer to Problem 82P

The radial and perpendicular component of electric field is (2kpcosθ/r3)_ and (kpsinθ/r3)_ respectively.

Explanation of Solution

Write the expression for the radial component of electric field.

Er=Vr        (V)

Here, Er is the radial component of electric field and (V/r) is the partial derivative of potential with respect to distance.

Write the equation for perpendicular component of electric field.

  Eθ=1rVθ        (VI)

Here, Eθ is the perpendicular component of electric field, (V/θ) is the partial derivative of potential with respect to angle.

Conclusion:

Substitute, kpcosθr2 for V in Equation (V) to find Er.

Er=[kpcosθr2]r=2kpcosθr3

Substitute, kpcosθr2 for V in Equation (VI) to find Eθ.

Eθ=1r(kpcosθr2)θ=kpsinθr3

Thus, the radial and perpendicular component of electric field is (2kpcosθ/r3)_ and (kpsinθ/r3)_ respectively.

(c)

To determine

The electric field at 90° and 0°.

(c)

Expert Solution
Check Mark

Answer to Problem 82P

The electric field at 90° and 0° are Eθ(90)=kpr3_.and Er(0)=2kpr3_

Explanation of Solution

Write the expression for the radial component of electric field.

Er=2kpcosθr3        (VII)

Write the expression for the perpendicular component of electric field.

Eθ=kpsinθr3        (VIII)

Conclusion:

For r>>a

Substitute, 90° for θ in Equation (VII) and (VIII) to find Er,Eθ.

  Er=2kpcos(90)r3=0

  Eθ=kpsin(90)r3=kpr3

For r>>a

Substitute, 0° for θ in Equation (VII) and (VIII) to find Er,Eθ.

  Er=2kpcos(0)r3=2kpr3

  Eθ=kpsin(0)r3=0

The results are reasonable since the component of electric field is having finite value.

Thus, the electric field at 90° and 0° are Eθ(90)=kpr3_.and Er(0)=2kpr3_.

(d)

To determine

The electric field at r=0.

(d)

Expert Solution
Check Mark

Answer to Problem 82P

The electric field at r=0 is Eθ(r=0)=_.and Er(r=0)=_

Explanation of Solution

Write the expression for the radial component of electric field.

Er=2kpcosθr3        (VII)

Write the expression for the perpendicular component of electric field.

Eθ=kpsinθr3        (VIII)

Conclusion:

Substitute, 0 for r in Equation (VII) and (VIII) to find Er,Eθ.

  Er=2kpcos(90)(0)3=

  Eθ=kpsin(90)(0)3=

The electric field at the centre of dipole is not infinite.

The results are not reasonable since the component of electric field is having infinite value.

Thus, The electric field at r=0 is Eθ(r=0)=_.and Er(r=0)=_.

(e)

To determine

The potential in Cartesian coordinate.

(e)

Expert Solution
Check Mark

Answer to Problem 82P

The potential in Cartesian coordinate is V=kpy(x2+y2)3/2.

Explanation of Solution

Write the expression for the potential.

V=kpcosθr2        (IX)

Conclusion:

Substitute, (x2+y2)1/2 for r, y(x2+y2)1/2 for cosθ in Equation (IX) to find V.

  V=kp[y(x2+y2)1/2](x2+y2)=kpy(x2+y2)3/2

Thus, the potential in Cartesian coordinate is V=kpy(x2+y2)3/2.

(f)

To determine

The x and y component of electric field.

(f)

Expert Solution
Check Mark

Answer to Problem 82P

The x and y component of electric field is (3kpxy/(x2+y2)5/2)_ and (kp(2y2x2)/(x2+y2)5/2)_ respectively.

Explanation of Solution

Write the expression for the x component of electric field.

Ex=Vx        (X)

Here, Ex is the x component of electric field and (V/x) is the partial derivative of potential with respect to x-distance.

Write the equation for y component of electric field.

  Ey=Vy        (XI)

Here, Ey is the y component of electric field, (V/y) is the partial derivative of potential with respect to y-distance.

Conclusion:

Substitute, kpy(x2+y2)3/2 for V in Equation (X) to find Ex.

Ex=(kpy(x2+y2)3/2)x=(3kpxy/(x2+y2)5/2)

Substitute, kpy(x2+y2)3/2 for V in Equation (XI) to find Ey.

Ey=(kpy(x2+y2)3/2)y=(kp(2y2x2)/(x2+y2)5/2)

Thus, the x and y component of electric field is (3kpxy/(x2+y2)5/2)_ and (kp(2y2x2)/(x2+y2)5/2)_ respectively.

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Chapter 20 Solutions

Principles of Physics: A Calculus-Based Text

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