The two charges in Fig.-1 are separated by a distance d = 100 cm, and Q = +5.00

Question

The two charges in Fig.-1 are separated by a distance d = 100 cm, and Q = +5.00 nC. Find the magnitude of the electric field at point A due to 2Q A) 4.50 times 10^5 V B) 4.50 times 10^5 J C) 4.50 times 10^5 N D) 4.50 times 10^5 N/C E) None Calculate the electric potential at point B due to Q. A) 3.18 times 10^3 V B) 3.18 times 10^3 V C) 3.18 times 10^3 V D)3.18 times 10^3 V E) None 9. What is the magnitude of the electric force between the two charges? A) 4.50 times 10^6 N B) 4.50 times 10^9 V C) 4.50 times 10^3 N D) 4.50 times 10^8 N E) None

Explanation / Answer

d = 1 cm , Q = 5 nC

7) E = kq/r^2

r = (1^2+1^2)^0.5 =1.414 cm

E = 9*10^9*2*5*10^-9/0.01414^2

E = 4.5*10^5 N/C

correct option is (D)

8) V = kq/r

V = 9*10^9*5*10^-9/0.01414

V = 3.18*10^3 V

9) F = kq1q2/r^2

F = (9*10^9*5*10^-9*2*5*10^-9)/0.01^2

F = 4.5*10^-3 N

correct option is (C)

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