Question

In many applications, oppositely charged parallel plates (with small holes cut for the beam to pass through) are used to accelerate beams of charged particles. In this figure, a proton is injected at $v_i$ into the space between the plates. The potential difference between the plates is $\Delta V$ (the left plate is at higher potential). What is the velocity of the proton as it exits the device?

• A. $v_f=\sqrt{{v_i}^2-\frac{2e\Delta V}{m}}$
• B. $v_f=\sqrt{{v_i}^2+\frac{e\Delta V}{m}}$
• C. $v_f=\sqrt{{v_i}^2+\frac{2e\Delta V}{m}}$
• D. $v_f=\sqrt{{v_i}^2-\frac{e\Delta V}{m}}$

Answer 1

The correct option is C $v_f=\sqrt{{v_i}^2+\frac{2e\Delta V}{m}}$

$v_f=\sqrt{v_1^2+\frac{e\Delta v}{m}}$

$E=\frac{1}{2}{mV}^2$

or, $e\Delta v=\frac{1}{2}{mv}_f^2$

or, $v_f=\sqrt{\frac{2e\Delta v}{m}}$

Since, $v_1=$ initial velocity

Hence, $v_f=\sqrt{v_1^2+\frac{2e\Delta v}{m}}$