Two large vertical and parallel metal plates having a separation of 1cm are connected to a DC voltage source of POTENTIAL DIFFERENCE X. A proton is released at rest midway between the two plates. It is found to move at 45° to the vertical just after release. Then X is nearly
a) l X 10^-5 V
b)1 X 10^-7V
c)1 X 10^-9V
d)1 X 10^-10V
Please explain the steps involved.

Asked by Renuka Ganesh | 20th May, 2014, 08:12: AM

Expert Answer:

The separation between the plates is given as d = 1 cm
Let E be the external electric field between the plates due to the applied potential diffrence X.
The proton released between the plates experiences a force due to the external electric field, Fx = q E
It experiences a force due its weight downwards, Fy = mg
Given that the proton is found to move at an angle 45° to the vertical just after the release.
We know that using the pararllelogram law of vector addition the direction of resultant vector can be determined using
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Putting α = 45 ° and θ = 90º ,
we can write Fx = Fy
i. e. q E = mg
As E = V / d
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X = 1 × 10-9 V
Correct option (C)

Answered by Jyothi Nair | 20th May, 2014, 10:18: AM

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