2 squareroot 2 kq/L 2 squareroot 2 kq^2/L^2 4kq^2/L^2 kq/L^2 kq^2/L^2 4kq/L^2 4k

Question

2 squareroot 2 kq/L 2 squareroot 2 kq^2/L^2 4kq^2/L^2 kq/L^2 kq^2/L^2 4kq/L^2 4kq^2/L A positive charge of 9.25 times 10^-5 C experiences a force of 0.291 N when located at a certain point. What is the electric field magnitude at that point? Answer in units of N/C. The diagram shows an isolated, positive charge Q, where point B is twice as far away from Q as point A. A charge of -3.79 mu C is located at the origin, and a charge of -2.94 mu C is located along the y axis at 2.91516 m. At what point along the y-axis is the electric field zero? The value of the Coulomb constant is 8.99 times 10^9 N middot m^2/C^2. Answer in units of m. The electric field near the surface of the earth points downward and has a magnitude of 160 N/C. Compare the upward electric force on an electron with the downward gravitational force: i.e., what is the ratio F_e/F_g? The elementary charge s 1.6 times 10^-19 C and the mass of an electron is 9.11 times 10^-31 kg. What is the magnitude of the charge that should be placed on a coin of mass 7.2 g so that the electric force balances the weight of the coin near the earth's surface? Answer in units of C. Two point charges at fixed locations produce an electric field as shown.

Explanation / Answer

10) Q = 9.25*10^-5 C

F = 0.291 N

As F = QE =====> E = F/Q = 0.291/9.25*10^-5 = 3145.95 N/C

12) As d2/d1=sqrt(q2/q1) ==> R/d1=1+sqrt(q2/q1) ==> d1=R/(1+sqrt(q2/q1))
where d1, d2 are distances from q1, q2 to null-field point, R is distance between q1 and q2

Then d1=R/(1+sqrt(q2/q1)) = 2.91516/[1 + sqrt(2.94/3.79)] = 1.5499963 = 1.55 m

13) Fe/Fg = Ee/mg = 160*1.6*10^-19/9.11*10^-31*9.81 = 2.86*10^12. The electric force is very much larger

14) q = mg/E = 7.2*10^-3*9.81/160 = 4.4145*10^-4 C

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