Ch16, Electric Charge and Electric Field P18.10 Two identical metal spheres attr

Question

Ch16, Electric Charge and Electric Field P18.10 Two identical metal spheres attract each other with a force of 5.00 x 10-6 N when they are 5.00 cm apart. The spheres are then touched together and then removed to the original separation where now a force of repulsion of 1.00 x 10-6 N is observed. What is the charge on each sphere after touching and before touching? Fr2 F'r2 after touch 2,-q1 +92 q2 =-= 2.78 × 10-19 C2 q=5.27 x 10-10 C 11 (29-91) =-5q2 -2q1-5q 0 solve via quadratic before touch q.-+3.45q-1.82x 10-9 C 92 -1.45,--7.64 x10-10 C P18.11 Two point charges repel each other with a force of 3.00 x 10-5 N when they are 20.0 cm apart. Find the force if the distance is reduced to 5.00 cm F-192 PseG)' (3.00 x 10-5 N)(5.00 ? 10-1 m)2s4 80x10-4 N (2.00x 10-1 m 2 5.00 × 10-2 m) -4.80 × 10-4 N

Explanation / Answer

Standard quadratic eqn has form as below,

ax2 + bx + c = 0

The solution of quadratic eqn can be calculated using Quadratic formula,

x= [-b +/- sqrt(b2-4ac)]/2a ------------(1)

Now we have eqn,

q12 - 2qq1 -5q2 = 0

Comparing this eqn with standard quadratic eqn we get,

a=1, b= -2q , c = -5q2

Plugging these values in eqn (1),

x = [-(-2q) +/- sqrt((-2q)2-4(1)( -5q2))]/2

x = [-(-2q) +/- sqrt((4q2)-4(1)( -5q2))]/2

x = [2q +/- sqrt(20q2+4q2)]/2

x = [2q +/- sqrt(24q2)]/2

x = [2q +/- (4.90q)]/2

x = [2q + (4.90q)]/2   or   x = [2q - (4.90q)]/2

x = (6.90/2)q    or    x = (-2.24/2)q

x = 3.45q    or    x = -1.45q

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