Question

If the electric field between the plates of a cathode ray oscilloscope be $1.2\times {10}^{4}N/C$ , the deflection that an electron will experience if it enters at right angles to the field with kinetic energy 2000 eV is (The deflection assembly is 1.5cm long) :

• A. $0.34cm$
• B. $3.4cm$
• C. $0.034mm$
• D. $0.34mm$

Answer 1

The correct option is D $0.34mm$

The electron moves against the electric field because it experiences a force due to the field.

The acceleration experienced by the electron in $y$ direction is $a_y=\frac{qE}{m}$

Thus, by kinematics relation $\Delta y=\frac{1}{2}{a}_y\Delta t^2$

As the electron moves with a constant speed in $x$ direction, $\Delta t=\frac{\Delta x}{v}$

The velocity of the electron in terms of the kinetic energy is $v=\sqrt{\frac{2KE}{m}}$

Using the above relations for $\Delta y$, we have $\Delta y=\frac{1}{2}\frac{qE}{m}(\frac{\Delta x}{v}{)}^2=\frac{E\Delta x^2}{4\times \frac{KE}{q}}$

Given $\frac{KE}{q}=2000\text{V}$ and $E=1.2\times {10}^4\text{N/C}$ and $\Delta x=1.5\times {10}^{-2}\text{cm}$

We get, $\Delta y=\frac{1.2\times {10}^4\times (1.5\times {10}^{-2}{)}^2}{4\times 2000}=0.3375\times {10}^{-3}\text{m}\approx 0.34\text{mm}$