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

Answer:

76.66V

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:

$U_3-U_1=-K_1\\q(V_3-V_1)=-\frac{mv_1^2}{2}\\U_3-U_2=-K_2\\q(V_3-V_2)=-\frac{mv_2^2}{2}$

Recall that $v_1=2v_2$, replacing:

$q(V_3-V_1)=-\frac{m(2v_2)^2}{2}\\q(V_3-V_1)=-\frac{4mv_2^2}{2}\\q(V_3-V_1)=-4K_2(1)\\q(V_3-V_2)=-K_2(2)$

Rewriting (1) for $-K_2$:

$q(V_3-V_1)=-4K_2\\-K_2=\frac{q(V_3-V_1)}{4}(3)$

Now, equaling (2) and (3):

$q(V_3-V_2)=\frac{q(V_3-V_1)}{4}$

Finally, rewriting for $V_3$

$(V_3-V_2)=\frac{V_3-V_1}{4}\\4V_3-4V_2-V_3=-V_1\\3V_3=4V_2-V_1\\V_3=\frac{4V_2-V_1}{3}\\V_3=\frac{4(115V)-231V}{3}\\V_3=76.66V$