Two small beads having charges q 1 and q 2 are fixed so that the centers of the

Question

Two small beads having charges q1 and q2 are fixed so that the centers of the two beads are separated by a distance d shown in the figure. A third small bead of charge q=10.0 x 10-9 C is placed between both charges a distance x from q1.

a) If q1 = 50.0 x 10-9 C, q2 = 24.0 x 10-9 C, d = 0.39 m and x =1/3 d, what force is exerted on q due to both charges? Answer in units of Newtons and use the direction from q1 to q2 as the positive direction.

b)  What force is exerted on q2 due to both charges q and q1?Answer in units of Newtons and use the direction from q1 to q2 as the positive direction.

c) If q1 = 50.0 x 10-9 C, q2 = - 24.0 x 10-9 C, d = 0.39 m and x =1/3 d, what force is exerted on q due to both charges? Answer in units of Newtons and use the direction from q1 to q2 as the positive direction.

Explanation / Answer

F = k q1 q2 / r2

a. F= k*50.0 x 10-9*10.0 x 10-9/0.132 - k* 24.0 x 10-9 *10.0 x 10-9/0.262

=2.343*10-4 N in positive x axis

b. F= k*50.0 x 10-9* 24.0 x 10-9 /0.392 + k*10.0 x 10-9* 24.0 x 10-9/0.262

F=1.03*10-4 N in positive x axis

c. F= k*50.0 x 10-9*10.0 x 10-9/0.132 + k* 24.0 x 10-9 *10.0 x 10-9/0.262

F= 2.98*10-4 N in positive x axis

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