A uniform electric field of 10NC^-1 is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically…

Given here, kinetic energy $\frac{1}{2}e=\frac{1}{2}m{u}_{x}^{2}\Rightarrow {u}_{x}=\sqrt{\frac{e}{m}}$21​e=21​mux2​⇒ux​=me​​

Acceleration of electron due to electric field is $a=\frac{eE}{m}$a=meE​, where, $e$e is charge on electron.

Using equation of motion, $s={u}_{x}t+\frac{1}{2}{a}_{x}{t}^{2}$s=ux​t+21​ax​t2 along $x$x direction, we get

$$L={u}_{x}t+0(\because {a}_{x}=0) \Rightarrow 10\times {10}^{-2}={u}_{x}t \Rightarrow t=\frac{{10}^{-1}}{{u}_{x}}....(1)$$L=ux​t+0(∵ax​=0)⇒10×10−2=ux​t⇒t=ux​10−1​....(1)

Using equation of motion, ${v}_{y}={u}_{y}+{a}_{y}t$vy​=uy​+ay​t along y direction, we get

$${v}_{y}=0+\frac{eE}{m}t=\frac{eE}{m}t...(2)$$vy​=0+meE​t=meE​t...(2)

From equation $(1)\mathrm{and}(2)$(1)and(2), we have

$${v}_{y}=\frac{eE}{m}(\frac{{10}^{-1}}{{u}_{x}})$$vy​=meE​(ux​10−1​)

Now, $\mathrm{tan}\theta =\frac{{v}_{y}}{{v}_{x}}=\frac{eE}{m}(\frac{{10}^{-1}}{{u}_{x}^{2}})=\frac{e\times 10\times {10}^{-1}}{m\times \frac{e}{m}}=1$tanθ=vx​vy​​=meE​(ux2​10−1​)=m×me​e×10×10−1​=1

The angle of deviation of the path of electron as it comes out of the field is $\theta =45^{\circ}$θ=45∘.