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Problems & Exercises

1.

(a) 1.25×10101.25×1010

(b) 3.13×10123.13×1012

3.

-600 C

5.

1.03 × 10 12 1.03 × 10 12

7.

9 . 09 × 10 13 9 . 09 × 10 13

9.

1.48×108C1.48×108C

11.

(a) 0.263 N

(b) If the charges are distributed over some area, there will be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force.

13.

The separation decreased by a factor of 5.

17.

F = k|q1q2|r2 = maa= kq2mr2 = (9.00×109N m2/C2) ( 1.60×10–19m)2 ( 1.67×10–27kg) ( 2.00×10–9m)2 = 3.45×1016m/s2 F = k|q1q2|r2 = maa= kq2mr2 = (9.00×109N m2/C2) ( 1.60×10–19m)2 ( 1.67×10–27kg) ( 2.00×10–9m)2 = 3.45×1016m/s2

18.

(a) 3.2

(b) If the distance increases by 3.2, then the force will decrease by a factor of 10 ; if the distance decreases by 3.2, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10.

20.

(a) 1.04×1091.04×109 C

(b) This charge is approximately 1 nC, which is consistent with the magnitude of charge typical for static electricity

23.

1 . 02 × 10 11 1 . 02 × 10 11

25.
  1. 0.859 m beyond negative charge on line connecting two charges
  2. 0.109 m from lesser charge on line connecting two charges
28.

8.75×1048.75×104 N

30.

(a) 6.94×108C6.94×108C

(b) 6.25 N/C6.25 N/C

32.

(a) 300 N/C (east)300 N/C (east)

(b) 4.80×1017 N (east)4.80×1017 N (east)

42.

(a) 2.12×105N/C2.12×105N/C

(b) one charge of +q+q

44.

(a) 0.252 N to the left

(b) x=6.07 cmx=6.07 cm

46.

(a)The electric field at the center of the square will be straight up, since qaqa and qbqb are positive and qcqc and qdqd are negative and all have the same magnitude.

(b) 2.04×107N/C(upward)2.04×107N/C(upward)

48.

0.102N,0.102N, in the yy direction

50.

(a) E=4.36×103N/C,35.0ºE=4.36×103N/C,35.0º, below the horizontal.

(b) No

52.

(a) 5.58×1011 N/C5.58×1011 N/C

(b)the coulomb force is extraordinarily stronger than gravity

54.

(a) 6.76×105 C6.76×105 C

(b) 2.63×1013m/s2(upward)2.63×1013m/s2(upward)

(c) 2.45×1018kg2.45×1018kg

56.

The charge q2q2 is 9 times greater than q1q1.

69.

(a) The forces are balanced in the x-direction, so the net force is vertical. It is composed of the sum of the vertical components of the Coulomb force and the gravitational force. F y =2 k q 1 q 2 r 2 +G m 1 m 2 r 2 cos45°= 2 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 8 +6.67× 10 11 (10.0)(10.0) 8 cos45°=4.36× 10 9  N F y =2 k q 1 q 2 r 2 +G m 1 m 2 r 2 cos45°= 2 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 8 +6.67× 10 11 (10.0)(10.0) 8 cos45°=4.36× 10 9  N

(b) No, it is in a metastable position. Since it cannot move horizontally, it cannot traverse any part of the track.

(c) F x = k q 1 q 2 4 +G m 1 m 2 4 1 2 + k q 1 q 2 12 +G m 1 m 2 12 3 2 = 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 4 1 2 6.67× 10 11 (10.0)(10.0) 4 1 2 + 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 12 3 2 6.67× 10 11 (10.0)(10.0) 12 3 2 N=2.60× 10 9  N F x = k q 1 q 2 4 +G m 1 m 2 4 1 2 + k q 1 q 2 12 +G m 1 m 2 12 3 2 = 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 4 1 2 6.67× 10 11 (10.0)(10.0) 4 1 2 + 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 12 3 2 6.67× 10 11 (10.0)(10.0) 12 3 2 N=2.60× 10 9  N F y = k q 1 q 2 4 +G m 1 m 2 4 3 2 k q 1 q 2 12 +G m 1 m 2 12 1 2 =2.08× 10 9  N 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 4 3 2 6.67× 10 11 (10.0)(10.0) 4 3 2 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 12 1 2 +6.67× 10 11 (10.0)(10.0) 12 1 2 N=6.36× 10 9  N F y = k q 1 q 2 4 +G m 1 m 2 4 3 2 k q 1 q 2 12 +G m 1 m 2 12 1 2 =2.08× 10 9  N 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 4 3 2 6.67× 10 11 (10.0)(10.0) 4 3 2 8.99× 10 9 (1.00× 10 9 )(2.00× 10 9 ) 12 1 2 +6.67× 10 11 (10.0)(10.0) 12 1 2 N=6.36× 10 9  N

(d) Yes.

(e) It will reach (1, 3) (1, 3) before it will change direction.

(f) The metastable positions where there is no component in one direction are (0,0) (0,0), (4,0) (4,0), (2,2) (2,2), and (2,2) (2,2). They number 4.

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