Two charges of 1.0 muC and - 2.5muC respectively.experience a force of attractio

Question

Two charges of 1.0 muC and - 2.5muC respectively.experience a force of attraction of 10 N. What is the separation between charges? A. 4.70 cm B. 29. cm C. 15.0 cm D. 19.8 cm E. 224.6 cm Two charges of -2.0muC, 1.0muC respectively are separated by a distance of 12 cm and held fixed in the position. The third charge of 3.0muC is placed along the line of two charges and moved around until the force it experiences is zero. Determine the distance between 1.0muCand3.0muCcharges. a. 24cm b.29 cm c.31cm d.25cm e.41cm Two charges of 1.0muC each are 1.2 cm apart The magnitude of the electric field midway between these charges is: a 5.0 times 10^4 N/C b.6.0 times 10^5 N/C c. 1.0 times 10^5 N/C d. 11.0 times 10^-5 N/C e.0.0 N/C Two charges. Q_1 = -2.0muC and Q_2 = 8.0muCare placed, as shown, on a 30degree -60degree right triangle of sides 1.0 cm and 1.73cm. The magnitude of the electric field at the held point FP, is: a. 1.85 times 10^4 N/C b.3.63 times 10^3 N/C c.3.12 times 10^8 N/C

Explanation / Answer

1) F = kq1q2/r^2 ========> r = sqrt(kq1q2/F)

r = sqrt[(9*10^9*1*2.5*10^-12)/1] = 0.15 m = 15 cm (Option C)

2) Let the distance between q1 and q3 be x meter. so,

kq1q3/x^2 = kq2q3/(0.12 + x)^2 ==========>  q1/x^2 = q2/(0.12 + x)^2

1/x^2 = 2/(0.12 + x)^2

(0.12 + x)^2 = 2x^2 =======> x^2 - 0.24x - 0.0144 = 0

x = 0.2897 m = 28.97 = 29 cm (option B)

3) E = kq/r^2

Here charges are of equal magnitude but E due to both charges is equal and opposite. So, net field will be zero at midpoint. (Option E is correct)

4) There is no figure.

  

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