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Question: A proton is released from rest, \(10cm\) from a charged sheet carrying charged density of \( - 2.21 ...

A proton is released from rest, 10cm10cm from a charged sheet carrying charged density of −2.21×10−9C/m2 - 2.21 \times {10^{ - 9}}C/{m^2}. It will strike the sheet after time (approximately)
(a). 4μs4\mu s
(b). 2μs2\mu s
(c). 22μs2\sqrt 2 \mu s
(d). 42μs4\sqrt 2 \mu s

Explanation

Solution

- Hint: The force that a charge experiences in the presence of another charge is known as electrostatic force. This force is also known as Coulomb force, after the name of its discoverer Charles-Augustin de Coulomb.

Complete step-by-step answer:
The charge experiences Coulomb’s force due to the formation of an electric field around the other charge. Imagine a fisherman entrapping a fish in its net and pulling it, this is somewhat how electric fields trap a charge.

We know that the magnitude of the electric field due to a charged sheet is given by the equation
E=σ2ε0E = \dfrac{\sigma }{{2{\varepsilon _0}}}
Here E=E = Electric field, σ=\sigma = Surface charge density and ε0={\varepsilon _0} = Permittivity of free space.
E=2.21×10−92×8.85×10−12E = \dfrac{{2.21 \times {{10}^{ - 9}}}}{{2 \times 8.85 \times {{10}^{ - 12}}}}
E=124.86N/CE = 124.86N/C
This electric field formed will exert an electrostatic force of attraction on the proton, which will be given by the equation
F=eEF = eE
Here, e=e = Charge of proton =1.6×10−19C = 1.6 \times {10^{ - 19}}C
F=1.6×10−19×124.86NF = 1.6 \times {10^{ - 19}} \times 124.86N (Equation 1)
We know by Newton’s second law that,
F=maF = ma
Here, m=1.67×10−27kg=m = 1.67 \times {10^{ - 27}}kg = Mass of proton and a=a = Acceleration of the proton
1.67×10−27×a=1.6×10−19×124.861.67 \times {10^{ - 27}} \times a = 1.6 \times {10^{ - 19}} \times 124.86
a=1.6×10−19×124.861.67×10−27a = \dfrac{{1.6 \times {{10}^{ - 19}} \times 124.86}}{{1.67 \times {{10}^{ - 27}}}}
a=1.2×1010m/s2a = 1.2 \times {10^{10}}m/{s^2}
Now, using the second equation of motion we know that
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2}
Here, s=s = distance =10cm=0.1m = 10cm = 0.1m , u=u = initial velocity of proton =0 = 0 (proton is at rest), t=t = time and a=a = Acceleration of the proton
So, 0.1=0×t+12×1.2×1010×t20.1 = 0 \times t + \dfrac{1}{2} \times 1.2 \times {10^{10}} \times {t^2}
t2=0.21.2×1010{t^2} = \dfrac{{0.2}}{{1.2 \times {{10}^{10}}}}
t=0.21.2×1010t = \sqrt {\dfrac{{0.2}}{{1.2 \times {{10}^{10}}}}}
t=4.08×10−6sec⁡∼4μst = 4.08 \times {10^{ - 6}}\sec \sim 4\mu s
Hence, option A is the correct choice.

Note: In the problem we are given a charged sheet carrying a charged density of −2.21×10−9C/m2 - 2.21 \times {10^{ - 9}}C/{m^2}. But while attempting the solution we mostly ignore the minus sign because the negative sign only represents the nature of the charge and this does not affect the magnitude of the electric field, the electrostatic force or time.