x 10\'N. b. 1.84 x 10N. C3.04 x 10\'N. 904x 10*N 108x10 N 0.57 kg and m2- 0.89 k
ID: 1882289 • Letter: X
Question
x 10'N. b. 1.84 x 10N. C3.04 x 10'N. 904x 10*N 108x10 N 0.57 kg and m2- 0.89 kg are located 18.6 um apart. What is the magnitude of the force that m, exerts on . . 72.1 x 10 N. d. 584 x 10' N. .39.5x10 N [6]#4. Point masses m.- m2? Answer:a 97.8 x 10 N. b. 86.4 x 10 N which represents two point charges For #5 and #6, use the figure at the right Q1--18.4 pC and Q'--4.99 C, which are separated by a distance of 172 cm. 0. [5]#5. Determine the force on Q2 due to Answer: a. 21.2 N, to the right. b. 27.9 N, to the right. c.33.4 N, to the right. d. 39.2 cm, to the left. .46.4 N, to the left. f [6]#6. Determine the location where the net electric field due to Q1 and Qs is zero? Answer: a. 5.89 cm, left of Q. b. 6.37 cm, left of Q. e.7.16 cm, left of Q. d.8 83 cm, let of 356 cm, right of Q [5] #7. Circle the TRUE answer choice or choices. a. There may be more than one correct answer. Two point charges that are separated by a small distance have the same amount of electric charge but are polarity. This collection of charges is called an electric dipole. b. Ifan electric dipole is placed in an electric field it will tend to rotate. c. Electric potential is a vector so it is importan t to correctly determine the component along each axis d. Electric field lines sometimes intersect to form 90° angles. Unlike the gravitational field the electric field is not a conservative field. e. [9] #8. The electric potential in a region of s field in the form pace is given byV-( 4.00 VI n) + (5.00 vm y -600v/me. D er inete E Ei+ Eyj + E,k at the point (O.0 m, +3.0 m, 1.0 m). Answers: a. E-(4.00 Vin)-(30.0 V/m)j+(18 V/m)k, c. E-(2.10 V/m)i - (31.2 V/m)j+ (0 V/m)k. .. E = (-7.00 V/m)-(2.81 V/m)j + (7.81 V/m)k b. E d. E- (3.10 V/m)i+(372 V/m)j -(0 V/m)k (3 (3.00 V/m)i (24.0 VIm)j+(O V/m)k f. 1 #9. At a certain location the electric field is given by E - (4.80 i +5.10j+7.70 k) V/m. Determine the electric flux throu face whose area vector at that location is A- -46.0 m2i. swer: a. 100 V.m b. 197 V.m c. 221 V.m d. 292 V.m e. 337 V.mfExplanation / Answer
4)
Force, F = G m1 m2/r^2
F = (0.57 x 0.89 x 6.67 x 10^-11)/(18.6 x 10^-6)^2 = 0.0978 N
Correct answer is option (a)
5)
F= kq1q2/r^2= 9*10^9*18.4*4.99*10^-12/0.172^2= 27.9 N
Option (b) is correct
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6.
Let d is the distance from q1 where net electric field due to both charges is zero, so at that point electric field due to q1 must balances electric field due to q2
K q1/d^2= k q2/(r-d) ^2
Sqrt (4.99/18.4) = (17.2/d)-1
d= 11.3 cm
From left of q2= 17.2-11.3= 5.89 cm
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7 option (a) (b) are correct.
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Comment in case any doubt . Goodluck
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