Question

A stationary block has a charge of +6.0×10−4 C. A 0.80-kg cart with a charge of +4.0×10−4 C is initially at rest separated by 4.0 m from the block. The cart is released and moves along a frictionless surface to a distance of 10.0 m from the block.

Consider the objects as point charges.

A) Determine the electric potential energy of the initial state.

B)

Determine the electric potential energy of the final state.

C)

Determine the change in electric potential energy.

D)

Determine the final speed of the cart. Assume that the change in electric potential energy is fully transformed into kinetic energy of the cart.

Answer 1

Answer:

A) U1 = 135.6 J

B) U2 = 21.7 J

C) -113.9 J

D) 16.8 m/s

Explanation:

The electric potential energy can be calculated by the following formula

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Since all the variables in the above formula are given, it is straightforward to calculate the electric potential energy in both cases.

A) $U_1 = \frac{1}{4\pi\epsilon_0}\frac{(6\times 10^{-4})(4\times 10^{-4})}{4^2} = 135.6~J$

B) $U_2 = \frac{1}{4\pi\epsilon_0}\frac{(6\times 10^{-4})(4\times 10^{-4})}{10^2} = 21.7~J$

C) The change in the electric potential energy is equal to the difference between U1 and U2.

Therefore,

$\Delta U = U_2 - U_1 = 21.7 - 135.6 = -113.9~J$

D) Since the change in the potential energy is fully converted into kinetic energy, therefore the change in the kinetic energy between both cases are equal to -113.9 J.

$\Delta U = -\Delta K = K_2 - K_1\\113.9 = K_2 - 0\\113.9 = \frac{1}{2}mv^2\\v = 16.8~{\rm m/s}$