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A small metal sphere, carrying a net charge of q1 = -2.80 µC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2 = -7.80 µC and mass 1.50 g, is projected toward q1. When the two spheres are 0.800 m apart, q2, is moving toward q1 with speed 22.0 m/s (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.

(a) What is the speed of q2 when the spheres are 0.400 m apart?

(b) How close does q2 get to q1?

Short Answer

Expert verified
  1. The speed of charge q2 when the spheres are 4.00 m apart is .
  2. The distance between the charges is 0.323 m.

Step by step solution

01

Charge at a distance

Charges at a distance apply force on each other, which is equal and opposite in direction. If the charges are of the same nature, this force will be repulsive; if not, it will be an attractive force. This force is known as coulomb force.

02

(a) Determination of the speed of charge q2 when the spheres are 4.00 m apart.

Let q2be 0.800 m away from q1 charge when it is at point a, and q1is at point b when it is 0.400 m away from q2. Refer to the image below.

Work is done by only the electric force in this case, So, Wother = 0.

According to the principle of conservation of energy,

Ka+Ua+Wother=Kb+Ub ….(i)

Here, the K- terms denote the kinetic energy, U-terms denote the potential energy given as,

U=14πε0q1q2r

Now, the value of kinetic energy of sphere a is,

Ka=12mva2=121.50×10-3kg22.0m/s2=0.3630J

The value of potential energy of sphere a is,

Ua=14πε0q1q2ra=8.988×109N.m2/C2-2.80×10-6C-7.80×10-6C0.800m=+0.2454J

Similarly, for sphere b,

Kb=12mvb2

And,

Ub=14πε0q1q2rb=8.988×109N.m2/C2-2.80×10-6C-7.80×10-6C0.400m=+0.4907J

From equation (i), Solving for Kb,

Kb=Ka+Ua-Ub12mvb2=+0.3630J+0.2454J-0.4907J=0.1177J

Thus, the speed of the charged particle q2 is,

vb=20.1177J1.50×10-3kg=12.5m/s

Thus, the speed of charge q2 when the spheres are 4.00 m apart is 12.5m/s.

03

(b) Determination of the distance between q2 and q1

Take a point c where the charge q2loses momentum, and its speed becomes zero. Refer to the image below and apply the Law of conservation of energy at points a and c,

The Law of conservation of energy at a and c is,

Ka+Ua=Uc …(ii)

Now,Ka=0.3630Jand Ua=+0.2454J from part (a).

At c, speed is zero. So, Kc is also zero.

The potential energy at c is,

Uc=14πε0q1q2rc

From equation (ii),

14πε0q1q2rc=+0.3630J+0.2454J=0.6084J

Solve for the distance,

rc=14πε0q1q20.6084=8.988×109N.m2/C2-2.80×10-6C-7.80×10-6C0.6084J=0.323m

Thus, the distance of the closest approach is 0.323 m.

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