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Question

An electric field of magnitude 1000 N/C is produced between two parallel plates having a separation of 2.0 cm as shown in the figure. If an electron is projected with speed 2.7×106 m/s at an angle of 60∘ with the field, the maximum height reached by the electron will be

[Assume gravity free space]

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Solution

The correct option is **C** 0.5 cm

We know that,

Charge of an electron, e=1.6×10−19 C

For the given situation,

Force on electron, F=eE (downwards)

The effective downward acceleration of electron is

geff=g−eEm

Assuming gravity free space

⇒geff=0−1.6×10−19×10009.1×10−31 (downward)

⇒geff=1.8×1014 m/s2

The maximum height achieved by the projectile is given by

H=u2sin2θ2geff

⇒H=(2.7×106)2×(sin30∘)22×1.8×1014=5×10−3 m

⇒H=0.5 cm

We know that,

Charge of an electron, e=1.6×10−19 C

For the given situation,

Force on electron, F=eE (downwards)

The effective downward acceleration of electron is

geff=g−eEm

Assuming gravity free space

⇒geff=0−1.6×10−19×10009.1×10−31 (downward)

⇒geff=1.8×1014 m/s2

The maximum height achieved by the projectile is given by

H=u2sin2θ2geff

⇒H=(2.7×106)2×(sin30∘)22×1.8×1014=5×10−3 m

⇒H=0.5 cm

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