Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 25, Problem 23P

(a)

To determine

The finite value(s) of x for which the electric field is zero.

(a)

Expert Solution
Check Mark

Answer to Problem 23P

4.83m is the point where the net electric field will be zero.

Explanation of Solution

The following figure shows the direction of the electric field at point P due to charges at points O and A; where, the point O is the origin.

Physics for Scientists and Engineers with Modern Physics, Technology Update, Chapter 25, Problem 23P , additional homework tip  1

Figure-(1)

Consider the distance between point O and P as r and distance between point O and A as x as shown in the Figure (1).

Write the expression for electric field.

    E=ke×qd2(i^) (I)

Here, E is the electric field, ke is the Coulomb’s constant, q is the charge, d is the distance from points and i^ is the field direction.

Substitute, +q for q and r for d in Equation (I) to calculate E1.

    E1=ke×qr2(i^)

Substitute 2q for q and rx for d Equation (I) to calculate E2.

    E2=ke×2q(rx)2(i^)

The net electric field at point P due to the charges +q and 2q is zero, therefore.

    Enet=E1+E2=0                                                                                                       (II)

Substitute ke×qd2(i^) for E1 and ke×2q(rx)2(i^) for E2 in Equation (II) to calculate Enet.

    ke×2q(rx)2(i^)+ke×qr2(i^)=02q(rx)2(i^)+qr2(i^)=02q(rx)2(i^)=qr2(i^)2q(rx)2=qr2

Further solve the above equation.

    2×r2=(rx)22×r2=(rx)2×r=(rx)(2×r)r=x

Further solve the above equation.

    r(21)=xr=x21                                                                                               (III)

Conclusion:

Substitute 2.00m for x in Equation (III) to calculate r.

    r=2.00m21=4.83m

Therefore, 4.83m is the point where, the net electric field will be zero.

(b)

To determine

The finite value(s) of x for which the electric potential is zero.

(b)

Expert Solution
Check Mark

Answer to Problem 23P

2m and 0.667m is the point where the net electric potential will be zero.

Explanation of Solution

The following figure shows the net electric field at point P due to the charges at points O and A; where, the point O is the origin.

Physics for Scientists and Engineers with Modern Physics, Technology Update, Chapter 25, Problem 23P , additional homework tip  2

Figure-(2)

Consider the distance between point O and P as r and the distance between points O and A as x as shown in figure (2).

Write the expression for the electric field.

    V=keqd (IV)

Here, V is the electric potential, ke is the Coulomb’s constant, q is the charge, d is the distance from points and i^ is the field direction.

Substitute, +q for q and r for d in Equation (I) to calculate V1.

    V1=ke×qr

Substitute 2q for q and rx for d Equation (I) to calculate V2.

    V2=ke×2q(rx)

The net electric potential at point P due to the charges +q and 2q is zero, therefore.

    Vnet=V1+V2=0                                                                                                            (V)

Substitute ke×qr for V1 and ke×2q(rx) for V2 in Equation (V) to calculate r.

    ke×2q(rx)+ke×qr=02q(rx)+qr=02q(rx)=qr2(rx)=1r

Further solve the above equation.

    2r=rxr=x                                                                                                             (VI)

Substitute 2q for q and xr for d Equation (I) to calculate V2.

    V2=ke×2q(xr)

Substitute ke×qr for V1 and ke×2q(xr) for V2 in Equation (II) to calculate r.

    ke×2q(xr)+ke×qr=02q(xr)+qr=02q(xr)=qr2(xr)=1r

Further solve the above equation.

    2r=xr3r=xr=x3                                                                                                               (VII)

Conclusion:

Substitute 2.00m for x in Equation (VI) to calculate r1.

    r1=2.00m

Substitute 2.00m for x in Equation (VI) to calculate r2.

    r2=2.00m3=0.667m

Therefore, 2m and 0.667m is the point where the net electric potential will be zero.

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Chapter 25 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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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