Question

Point P is at a potential of 300.3 kV, and point S is at a potential of 191.3 kV. The space between these points is evacuated. When a charge of +2e moves from P to S, by how much does its kinetic energy change?

Answer 1

Answer:

Change in kinetic energy = ∆KE=3.584 ×10^-14 J

Explanation:

Charge = q = 2e

Change in voltage=∆V= 191.3 – 303.3 = -112KV

We know that:  

Change in potential energy = ∆U = q∆V

As,

Change in kinetic energy= ∆KE = - ∆U = - q∆V

                                            ∆KE=-2e(-112K)

                                            ∆KE=3.584 ×10^-14 J

Hence, the change in kinetic energy will be 3.584 ×10^-14 J.

Answer 2

Answer:

$3.49*10^{-14}J$

Explanation:

The potential energy is expressed as

$W=-q*potiential$

also the total workdone in a system is the sum of the potential energy and kinetic energy, but in most cases the potential energy can be zero.

we can write

$K.E=W=-q * change in potential$

the change in potential is express as

$change in potential = V_{2}-V_{1}$

Hence we can calculate the kinetic energy change as

$K.E=-2(1.6*10^{-19})(191.3-300.3)*10^{3}\\K.E=3.49*10^{-14}J$