A uniform electric field of 10 N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle () of deviation of the path of electron as it comes out of the field is ___ (in degree
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol.
$$
\begin{aligned}
& 0.5 e=\frac{1}{2} m v_x^2 \Rightarrow v_x=\sqrt{\frac{e}{m}} \\
& \text { Along } x L=v_x t=\sqrt{\frac{e}{m}} t \\
& \text { Along } y v_y=\frac{e}{m} t \\
& \text { dividing } \frac{v_y}{L}=E_{\sqrt{ }} \sqrt{\frac{e}{m}}=E v_x \\
& \Rightarrow \tan \theta=\frac{v_y}{v_x}=E \times L=10 \times 0.1=L \\
& \qquad \theta=45^{\circ}
\end{aligned}
$$
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇