Velocity-Selector Capacitor Voltage for a Moving Electron: 04-BS-9, May 2014, Q8

Question 8: Velocity-Selector Capacitor Voltage for a Moving Electron (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Electron speed $v$ (West)	$1\times10^5$ m/s
Magnetic field $B$ (vertical, Down)	$0.5$ T
Plate separation $d$	$1$ mm $=1\times10^{-3}$ m

Find. The capacitor voltage, plate polarity, and plate orientation needed so the electric force exactly cancels the magnetic force on the electron.

Electron moving West in a downward B field. The required E points North, so the South-facing plate is positive.

Approach. Compute the magnetic force on the electron with $\vec F=q\vec v\times\vec B$ (careful with the electron's negative charge), then find the electric field that produces an equal and opposite force, and finally the plate voltage and polarity that produce that field.

Magnetic force direction and magnitude. Take East$=\hat x$, North$=\hat y$, Up$=\hat z$, so $\vec v=-v\hat x$ (West) and $\vec B=-B\hat z$ (Down). For the electron ($q=-e$): $$\vec F_B=q(\vec v\times\vec B)=(-e)\big[(-v\hat x)\times(-B\hat z)\big]=(-e)(vB)(\hat x\times\hat z)=(-e)(vB)(-\hat y)=evB\,\hat y$$ so $\vec F_B$ points North, magnitude $$F_B=evB=(1.6\times10^{-19})(10^5)(0.5)$$ $$F_B=\boxed{8.0\times10^{-15}\ \text{N, North}}$$
Required electric field. Cancellation needs $\vec F_E=-\vec F_B$ (South) on the electron. Since $\vec F_E=q\vec E=-e\vec E$, a electron force to the South requires $\vec E$ pointing North: $$E=\frac{F_B}{e}=vB=(10^5)(0.5)$$ $$E=\boxed{5.0\times10^{4}\ \text{V/m, North}}$$
Plate voltage, polarity and orientation. A uniform field points from the $+$ plate to the $-$ plate, so a North-pointing $E$ needs the South plate positive and the North plate negative, with both plates vertical and facing North–South (perpendicular to the required field): $$V=Ed=(5.0\times10^4)(1\times10^{-3})$$ $$V=\boxed{50\ \text{V, South plate positive, North plate negative}}$$

Quantity	Result
Magnetic force on electron	$8.0\times10^{-15}$ N, North
Required $E$	$5.0\times10^4$ V/m, North
Capacitor voltage $V$	$50$ V; South plate $+$, North plate $-$; plates vertical, facing N–S

Concept & Theory

Topic. Velocity selector — balancing the Lorentz magnetic and electric forces on a charged particle.

Key relations. $\vec F=q(\vec E+\vec v\times\vec B)$; balance condition $\vec E=-\vec v\times\vec B$ (for the force to vanish); uniform-field capacitor $E=V/d$.

Why this works. This is the standard velocity-selector configuration (as in a mass spectrometer): the magnetic force on a moving charge is always perpendicular to its velocity, so an electric field perpendicular to that same direction, of the right magnitude and sign, can exactly cancel it for one specific speed.

Common pitfall. Forgetting the electron's negative charge when applying $\vec F=q\vec v\times\vec B$ — using $+e$ throughout would flip every direction in the answer (magnetic force South instead of North, capacitor polarity reversed).

Source. Sadiku, Elements of Electromagnetics, Ch. 8 (Forces on moving charges, Lorentz force); Young & Freedman, University Physics, Ch. 27 (Velocity selector / mass spectrometer).

Question 8 of 8: Velocity-Selector Capacitor Voltage for a Moving Electron

Question 8: Velocity-Selector Capacitor Voltage for a Moving Electron (20 marks)