The figure shows an arrangement of two -4.5 nC charges,each separated by 5.0 mm from a proton.If the two negative charges are held fixed at their locations and the proton is given an initial velocity v as shown in the figure,what is the minimum initial speed v that the proton needs to totally escape from the negative charges? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C², e = 1.60 × 10^-19 C,m_proton= 1.67 × 10^-27 kg)

Question

Quizplus · Accepted Answer

The potential energy of the proton in the electric field created by the two negative charges is given by U=qV, where q is the charge of the proton and V is the electric potential at the location of the proton. The electric potential can be calculated as V=kq1/r1+kq2/r2, where k is the Coulomb constant, q1 and q2 are the charges of the negative charges and r1 and r2 are the distances between the negative charges and the proton. At the initial position of the proton, r1=5.0mm and r2=5.0mm. Thus, V=k(-4.5nC)/5.0mm+k(-4.5nC)/5.0mm+k(1.6e-19C)/5.0mm= -1.95 × 10^7 V. The minimum initial kinetic energy required for the proton to escape from the electric field is equal to its initial potential energy, so we have the equation (1/2)mv^2=qV, where m is the mass of the proton and v is its initial speed. Substituting the values, we get v=sqrt(2qV/m)=1.8 × 10^6 m/s, which is choice A.

The Figure Shows an Arrangement of Two -4