Question

A charge of 22.0 uC is held fixed at the origin.

Part A. If a -7.00 uC charge with a mass of 3.50 g is released from rest at the position (0.925 m, 1.17 m), what is its speed when it is halfway to the origin?

v = m/s

Part B. Suppose the -7.00 uC charge is released from rest at the point x = 1/2(0.925m) and y = 1/2(1.17m). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A?

a. Greater than the speed found in part A

b. Less than the speed found part in A

c. Equal to the speed found in part A

3. Find the speed of the charge for the situation described in part B.

v = m/s

Answer 1

Part A - Calculating initial potential energy of the system

V = kQ₁Q₂ / r

where r₁ = = 1.49m

and k = 9*10⁹

So V₁ = -0.93 J

When its halfway to the origin then r₂ = 1.49/2 = 0.745m

So V₂ = -1.86 J

So change in potential energy = 0.93J

Now this change in potential energy is transformed to kinetic energy of the charge

So 0.93 = mv²/2

which gives v = 23.05m/s

Part B - a) Greater than the speed found in part A

3) From above formula r₁ = 0.745 we get initial potential energy as

V₁ = -1.86 J

Now when charge is half way t the origin then r₂ = 0.3725m

So V₂ = -3.72 J

So change in potential energy = 1.86 J

So 1.86 = mv²/2

which gives v = 32.6m/s