Chapter 22, Problem 41P

A 5.0-g object carries 3.8 μC. It acquires speed v when accelerated from rest through a potential difference V. If a 2.0-g object acquires twice the speed under the same circumstances, what’s its charge?

41. INTERPRET The problem asks for the charge of the particle which has been accelerated through a potential difference. We can find the magnitude of this potential difference given the information of the speed acquired upon traversing the potential difference by the mass with the given charge.

Develop The speed acquired by a charge q, starting from rest at point A and moving through a potential difference of V can be found using the work-energy theorem. The result is

ΔK_AB =q ΔV_AB $\Rightarrow \frac{1}{2}m{v}^2=q\Delta V_{AB}\Rightarrow v=\sqrt{\frac{2qq\Delta V_{AB}}{m}}$

This is the work-energy theorem for the electric force. A positive charge is accelerated in the direction of decreasing potential (i.e., increasing electric field). If we have two masses moving through the same potential difference, the ratio of their speeds would be

$\frac{v_2}{v_1}=\sqrt{\frac{2q_{2}V/m_2}{2q_{1}V/m1}}=\sqrt{\frac{q_2}{q_1}\frac{m_1}{m_2}}$

Evaluate If the second object acquires twice the speed of the first object (v₂/v₁ =2), moving through the same potential difference we find its charge from the equation above to be

$q_2=\left(\frac{m_2}{m_1}\right){\left(\frac{v_2}{v_1}\right)}^2{q}_1=\left(\frac{2g}{5g}\right){\left(2\right)}^2\left(3.8\mu C\right)=6.1\mu C$

ASSESS The speed of the particle moving through a potential difference is proportional to the square root of its charge, and inversely proportional to the square root of its mass.