Electrostatics Question 4
Question 4 - 25 January - Shift 1
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 $(\theta)$ of deviation of the path of electron as it comes out of the field is (in degree).
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Answer: (45)
Solution:
Formula: Electric force
$0.5 e=\frac{1}{2} mv_x^{2} \Rightarrow v_x=\sqrt{\frac{e}{m}}$
Along $x=v_x t=\sqrt{\frac{e}{m}} t$
Along y $v_y=\frac{e E}{m} t$
$dividing \frac{v_y}{L}=E \sqrt{\frac{e}{m}}=E v_x$
$\Rightarrow Tan \theta=\frac{v_y}{v_x}=E \times L=10 \times 0.1=1$
$\theta=45^{\circ}$
