Question 445: Electrostats & Capacitance

Question

A uniform electric field $$\mathrm{E}=(8 \mathrm{~m} / \mathrm{e}) \,\mathrm{V} / \mathrm{m}$$ is created between two parallel plates of length $$1 \mathrm{~m}$$ as shown in figure, (where $$\mathrm{m}=$$ mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of $$2 \mathrm{~m} / \mathrm{s}$$. The angle of the deviation $$(	heta)$$ of the path of the electron as it comes out of the field will be _________.

Accepted Answer

$$E = {{8\,m} \over e}$$ V/m
$$l = 1$$ m
$${v_x} = 2$$ m/s
$${a_y} =  - 8$$ m/s
2
$$t = {l \over {{v_x}}} = {1 \over 2}s$$
$$ \Rightarrow |{v_y}| = 4$$ m/s
$$\Rightarrow$$ angle of deviation = $$	heta$$
$$	an 	heta  = {{{v_y}} \over {{v_x}}}$$
$$	heta  = {	an ^{ - 1}}\left( {{4 \over 2}} ight) = {	an ^{ - 1}}(2)$$

Answer

$$
	an ^{-1}(4)
$$

Answer

$$
	an ^{-1}(2)
$$

Answer

$$	an ^{-1}\left(\frac{1}{3}ight)$$

Answer

$$	an ^{-1}(3)$$

Electrostats & Capacitance

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