Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 25, Problem 17P

(a)

To determine

The electric potential of the configuration of the three fixed charges.

(a)

Expert Solution
Check Mark

Answer to Problem 17P

The potential energy of the configuration is 45.0μJ.

Explanation of Solution

The following figure shows the placement of charges q1, q2 and q3. A fourth charge q4 is placed at point P.

Physics for Scientists and Engineers With Modern Physics, Chapter 25, Problem 17P

Figure-(1)

Write the expression for the configuration of the three fixed point’s electric potential energy.

    U=ke[q1×q2r12+q1×q3r13+q2×q3r23]                                                                             (I)

Here, U is the electric potential of three charges, ke is the Coulomb constant, q1, q2 and q3 are charges and r12, r13 and r23 are the distance between the charges.

Conclusion:

Substitute 9.0×109Nm2/C2 for ke, 20.0nC for q1, 20.0nC for q2, 10nC for q3, 4.00cm for r13 and r23, 8.00cm for r12 in Equation (I) to calculate U.

    U=(9.0×109Nm2/C2)[(20.0×20.0)nC28.00cm+(10.0×20.0)nC24.00cm+(20.0×10.0)nC24.00cm]=9.0×109Nm2/C2[400nC28.00cm+200nC24.00cm+200nC24.00cm]=9.0×109Nm2/C2[(400nC2)(1C109nC)2(8.00cm)(1m100cm)]=9.0×109Nm2/C2[4×1016C20.08m]

Further solve the above equation.

    U=(45×106Nm)(1J1Nm)=45×106J(1μJ106J)=45μJ

Therefore, the potential energy of the configuration is 45μJ.

(b)

To determine

The speed of the fourth particle, after it has moved freely to a very large distance away from its point.

(b)

Expert Solution
Check Mark

Answer to Problem 17P

The particle’s speed is 34.6km/s.

Explanation of Solution

Calculate the distance between the charges, q1 and q4 and the distance between the charges, q2 and q4 in Figure-(1).

Write the expression for Pythagoras theorem.

    r=r12+r22                                                                                                            (II)

Here, r is the hypotenuse side length, r1 is the adjacent side length and r2 is the opposite side length.

Write the expression for net potential due to the three charges at point P.

    V=ke[q1r14+q2r24+q3r34]                                                                                            (III)

Here, V is the net potential due to the three charges at point P.

Write expression when kinetic energy is converted into potential energy in the equilibrium position.

    12×m×v2=q4×VP

Here, v is the particle’s speed, m is the particle’s mass, q4 is the particle’s charge and VP is the particle’s potential at the point P.

Rewrite the above equation for v.

    12×m×v2=q4×VPv2=q4×VP12×mv=q4×VP12×mv=2×q4×VPm                                                                                         (IV)

Conclusion:

Consider triangle q1q3q4 and triangle q2q3q4. in Figure (1).

Substitute 4.00cm for r2 and 3.00cm for r1 in Equation (II) to calculate r14 and r24.

    r14=r24=(4.00cm)2+(3.00cm)2=(16+9)cm2=25cm2

Further solve the above equation.

  r14=r24=5cm(1m100cm)=0.05m

Substitute 9.0×109Nm2/C2 for ke, 20.0nC for q1, 20.0nC for q2, 10.0nC for q3, 0.05m for r14 and r24, 3.0cm for r34 in Equation (III) to calculate VP.

    VP=9.0×109Nm2/C2(20.0nC0.05m+(20.0nC)0.05m+10.0nC3cm(102m1cm))=9.0×109Nm2/C2(200.05m+(20)0.05m+100.03m)nC(109C1nC)=9.0×109Nm2/C2(333.33)×109C/m=3×103V

Substitute 40.0×109C for q4, 2.00×1013kg for m and 3×103V for VP in equation (IV) to calculate v.

    v=2(40.0×109C)×(3×103V)2.00×1013kg=240×107CV2kg=120×107CV/kg=34.6×103m/s(1km/s103m/s)

Further solve the above equation.

    v=34.6km/s

Therefore, the particle’ speed is 34.6km/s.

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Chapter 25 Solutions

Physics for Scientists and Engineers With Modern Physics

Ch. 25 - Prob. 7OQCh. 25 - Prob. 8OQCh. 25 - Prob. 9OQCh. 25 - Prob. 10OQCh. 25 - Prob. 11OQCh. 25 - Prob. 12OQCh. 25 - Prob. 13OQCh. 25 - Prob. 14OQCh. 25 - Prob. 15OQCh. 25 - Prob. 1CQCh. 25 - Prob. 2CQCh. 25 - When charged particles are separated by an...Ch. 25 - Prob. 4CQCh. 25 - Prob. 5CQCh. 25 - Prob. 6CQCh. 25 - Oppositely charged parallel plates are separated...Ch. 25 - Prob. 2PCh. 25 - Prob. 3PCh. 25 - How much work is done (by a battery, generator, or...Ch. 25 - Prob. 5PCh. 25 - Starting with the definition of work, prove that...Ch. 25 - Prob. 7PCh. 25 - (a) Find the electric potential difference Ve...Ch. 25 - Prob. 9PCh. 25 - Prob. 10PCh. 25 - Prob. 11PCh. 25 - Prob. 12PCh. 25 - Prob. 13PCh. 25 - Prob. 14PCh. 25 - Prob. 15PCh. 25 - Two point charges Q1 = +5.00 nC and Q2 = 3.00 nC...Ch. 25 - Prob. 17PCh. 25 - Prob. 18PCh. 25 - Given two particles with 2.00-C charges as shown...Ch. 25 - Prob. 20PCh. 25 - Four point charges each having charge Q are...Ch. 25 - Prob. 22PCh. 25 - Prob. 23PCh. 25 - Show that the amount of work required to assemble...Ch. 25 - Prob. 25PCh. 25 - Prob. 26PCh. 25 - Prob. 27PCh. 25 - Prob. 28PCh. 25 - Prob. 29PCh. 25 - Prob. 30PCh. 25 - Prob. 31PCh. 25 - Prob. 32PCh. 25 - How much work is required to assemble eight...Ch. 25 - Four identical particles, each having charge q and...Ch. 25 - Prob. 35PCh. 25 - Prob. 36PCh. 25 - Prob. 37PCh. 25 - Prob. 38PCh. 25 - Prob. 39PCh. 25 - Prob. 40PCh. 25 - Prob. 41PCh. 25 - Prob. 42PCh. 25 - Prob. 43PCh. 25 - Prob. 44PCh. 25 - Prob. 45PCh. 25 - Prob. 46PCh. 25 - Prob. 47PCh. 25 - The electric field magnitude on the surface of an...Ch. 25 - Prob. 49PCh. 25 - Prob. 50PCh. 25 - Prob. 51PCh. 25 - Prob. 52PCh. 25 - Prob. 53APCh. 25 - Prob. 54APCh. 25 - Prob. 55APCh. 25 - Prob. 56APCh. 25 - Prob. 57APCh. 25 - Prob. 58APCh. 25 - Prob. 59APCh. 25 - Prob. 60APCh. 25 - Prob. 61APCh. 25 - Prob. 62APCh. 25 - Prob. 63APCh. 25 - Prob. 64APCh. 25 - Prob. 65APCh. 25 - Prob. 66APCh. 25 - Prob. 67APCh. 25 - Prob. 68APCh. 25 - Review. Two parallel plates having charges of...Ch. 25 - When an uncharged conducting sphere of radius a is...Ch. 25 - Prob. 71CPCh. 25 - Prob. 72CPCh. 25 - Prob. 73CPCh. 25 - Prob. 74CPCh. 25 - Prob. 75CPCh. 25 - Prob. 76CPCh. 25 - Prob. 77CP
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