Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 17, Problem 31P

(a)

To determine

The electric potential at the midpoint of two point charges.

(a)

Expert Solution
Check Mark

Answer to Problem 31P

The electric potential at the midpoint of two point charges is 7.35×104V.

Explanation of Solution

The charges are 35.0 nC and 55.0 nC. The first charge is placed at the origin and the second charge is placed at a distance 2.20cm from the origin.

Physics, Chapter 17, Problem 31P

Write the expression for potential at point due to two charges

    Va=k(q1r1+q2r2)                                                                                         (I)

Here, Va is the electric potential at the midpoint of the two charges, k is the Coulomb’s constant, q1, q2 are the charges and r1r2  are their corresponding distance between the charges and the midpoint of the two charges.

Write the expression for midpoint

  M=x1+x22                                                                                                   (II)

Here, M is the midpoint of the two charges, x1 is the position of the first charge and x2 is the position of the second charge.

Substitute 0 for x1 and 2.2cm for x2 in (II) to find M

  M=(0+2.20)cm2=1.1 cm

The two charges are located on either side of the midpoint. Thus the distances r1 and r2 are same.

Substitute 8.988×109Nm2C2 for k35.0 nC for q1, 55.0 nC for q2, 1.1 cm for r1 and 1.1 cm for r2 in (I) to find Va. Here, Va is the potential at the midpoint.

  Va=8.988×109Nm2C2(35.0 nC1.1 cm+55.0 nC1.1 cm)=8.988×109Nm2C2(35.0 C1.1 m+35.0 C1.1 m)×109102=7.35×104V

Thus, the electric potential at the midpoint is 7.35×104V.

(b)

To determine

The electric potential at a point +3.40cm from the origin.

(b)

Expert Solution
Check Mark

Answer to Problem 31P

The electric potential at a distance +3.40cm from the origin is 5.04×104V.

Explanation of Solution

The first charge is at a distance +3.40cm from the point and the second charge is at a distance 3.40cm2.20cm from the point. (Refer figure 1)

Substitute 8.988×109Nm2C2 for k35.0 nC for q1, 55.0 nC for q2, +3.40cm for r1 and 3.40cm2.20cm for r2 in (I) to find Vb. Here, Va is the potential at a distance +3.40cm.

  Vb=8.988×109Nm2C2(35.0 nC3.40 cm+55.0 nC3.40cm2.20cm)=8.988×109Nm2C2(35.0 C3.40 m+35.0 C1.2 m)×109102=5.04×104V

Thus, the electric potential at a distance +3.40cm from the origin is 5.04×104V.

(c)

To determine

The work done by an external force in moving a charge at a distance +3.40cm from the origin

to the midpoint of the two charges.

(c)

Expert Solution
Check Mark

Answer to Problem 31P

The work done is 1.04mJ.

Explanation of Solution

The charge being moved is 45.0nC.

Write the expression for work done by an external agent in moving a charge

  W=ΔU                                                                                                  (III)

Here, W is the work done by the external force and ΔU is the change in electric potential in moving the charge from b to a.

Write the expression for ΔU

  ΔU=q(VfVi)                                                                                         (IV)

Here, q is the charge being moved Vi is the potential at a distance +3.40cm from the origin and  Vf is the potential at the midpoint of the two charges.

Equate (IV) and (III)

  W=q(VfVi)                                                                                             (V)

Substitute  45.0nC for q5.04×104V for Vi and 7.35×104V for Vf in (V) to find W

    W=45.0nC[(7.35×104V)(5.04×104V)]=+2.00×109C(899V1348V)=1.04×103J=1.04mJ

Thus, the work done is 1.04mJ.

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Chapter 17 Solutions

Physics

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