In Vuille and Serway\'s College Physics on page 510, Exercise 15.5 in the middle

Question

In Vuille and Serway's College Physics on page 510, Exercise 15.5 in the middle of the page. The problem is an extention of Example 15.5. The answer is given but I am having trouble figuring out how to get the answer.

Three electric charges form a right triangle. q1 is at the origin, -q2 is 0.3m to the right of q1 on the x-axis, and -q3 is 0.4m above q1 on the y-axis and 0.5m from -q2 on the diagonal (the hypotenuse of the triangle).

q1= 7 x 10^-6 C

q2= -5 x 10^-6 C

q3= -7 x 10^-6 C



A. What is the magnitude and direction of the electric field at q2 due to q1 and q3?

B. What is the magnitude and direction of the force on q2?

Explanation / Answer

due to q1. F12=kq1q2/0.3^2=3.5(N). F12 is attractive, so its direction (with q2) is negative x-axis. due to q3. F23=kq2q3/0.5^2=1.26(N). F23 is repulsive, so let alpha be the angle F23 make with the x-axis. tan alpha = -0.4/0.3 so alpha=-53.1(deg). ------ in total. Fx=-3.5+1.26*cos53.1=-2.7. Fy=1.26*sin(-53.1)=-1 so F=sqrt(Fx^2+Fy^2)=2.9. tan alpha = 1/2.7 so alpha=200(deg)

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