Question

Two protons are released from rest, one from location 1 and another from location 2. When these two protons reach location 3, the first proton has a speed that is 2 times the speed of the second proton. If the electric potentials at locations 1 and 2 are 231 V and 115 V, respectively, what is the electric potential at location 3?

Answer 1

76.66V

Step-by-step explanation:

Accorging to law of conservation of energy:

`\Delta E=0\\E_f=E_i\\K_f+U_f=K_i+U_i`}

Since the protons are released from rest, the initial kinetic energy is zero and the electric potencial energy is given by:
`U=qV`

Now from
`U_f-U_i=-K_f`, we have:

`U_3-U_1=-K_1\\q(V_3-V_1)=-(mv_1^2)/(2)\\U_3-U_2=-K_2\\q(V_3-V_2)=-(mv_2^2)/(2)\\`

Recall that
`v_1=2v_2`, replacing:

`q(V_3-V_1)=-(m(2v_2)^2)/(2)\\q(V_3-V_1)=-(4mv_2^2)/(2)\\q(V_3-V_1)=-4K_2(1)\\q(V_3-V_2)=-K_2(2)`

Rewriting (1) for
`-K_2`:

`q(V_3-V_1)=-4K_2\\-K_2=(q(V_3-V_1))/(4)(3)\\`

Now, equaling (2) and (3):

`q(V_3-V_2)=(q(V_3-V_1))/(4)`

Finally, rewriting for
`V_3`

`(V_3-V_2)=(V_3-V_1)/(4)\\4V_3-4V_2-V_3=-V_1\\3V_3=4V_2-V_1\\V_3=(4V_2-V_1)/(3)\\V_3=(4(115V)-231V)/(3)\\V_3=76.66V`