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Question: A proton is projected with a kinetic energy of \(18eV\) from infinity towards a fixed positive point...

A proton is projected with a kinetic energy of 18eV18eV from infinity towards a fixed positive point charge 20pC20pC. The smallest distance of proton from the fixed charge is
(A) 1cm1cm
(B) 4cm4cm
(C) 6cm6cm
(D) 9cm9cm

Explanation

Solution

Hint
The kinetic energy of the proton will be converted to the potential energy at the smallest distance to the point charge. So using the law of conservation of energy, we can find the smallest distance of approach.
In this solution, we will be using the following formula,
⇒U=kq1q2r\Rightarrow U = \dfrac{{k{q_1}{q_2}}}{r}
where UU is the potential energy
⇒k=14πεo\Rightarrow k = \dfrac{1}{{4\pi {\varepsilon _o}}} where εo{\varepsilon _o} is the permittivity in free space.
q1{q_1} and q2{q_2} are the charges.
and rr is the distance between them.

Complete step by step answer
Then the proton approaches the charge, the kinetic energy of the proton changes to its potential energy and at the distance of closest approach, the whole kinetic energy changes to potential energy.
Now the kinetic energy of the proton is given in the question as,
K.E.=18eVK.E. = 18eV. Now we can change the kinetic energy unit from electron-volt to volt by multiplying it with the electronic charge given by, 1e=1.6×10−19C1e = 1.6 \times {10^{ - 19}}C.
Therefore we get the kinetic energy of the electron as,
⇒K.E.=18×1.6×10−19V\Rightarrow K.E. = 18 \times 1.6 \times {10^{ - 19}}V
Now the potential energy of the proton at the closest distance rr is given by,
⇒U=kq1q2r\Rightarrow U = \dfrac{{k{q_1}{q_2}}}{r}
Where k=14πεo=9×109Nm2/C2k = \dfrac{1}{{4\pi {\varepsilon _o}}} = 9 \times {10^9}N{m^2}/{C^2}
In the question we are provided that the charge the proton approaches is 20pC20pC
So q1{q_1} is the charge of the proton given by, q1=1.6×10−19C{q_1} = 1.6 \times {10^{ - 19}}C
And q2{q_2} is the charge provided as, q2=20pC=20×10−12C{q_2} = 20pC = 20 \times {10^{ - 12}}C
So we can write the potential energy as, U=9×109×1.6×10−19×20×10−12rU = \dfrac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 20 \times {{10}^{ - 12}}}}{r}
Now from the law of conservation of energy,
⇒K.E.=U\Rightarrow K.E. = U
So substituting the values of the kinetic and potential energies we get
⇒18×1.6×10−19=9×109×1.6×10−19×20×10−12r\Rightarrow 18 \times 1.6 \times {10^{ - 19}} = \dfrac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 20 \times {{10}^{ - 12}}}}{r}
We can cancel the value of 1.6×10−19C1.6 \times {10^{ - 19}}C from both the sides of the equation and we get,
⇒18=9×109×20×10−12r\Rightarrow 18 = \dfrac{{9 \times {{10}^9} \times 20 \times {{10}^{ - 12}}}}{r}
Now we take the rr from the denominator of the RHS to the LHS and get,
⇒r=9×109×20×10−1218\Rightarrow r = \dfrac{{9 \times {{10}^9} \times 20 \times {{10}^{ - 12}}}}{{18}}
On calculating the numerator we have,
⇒r=0.1818\Rightarrow r = \dfrac{{0.18}}{{18}}
On doing the division we get the value of the distance of closest approach as,
⇒r=0.01m\Rightarrow r = 0.01m
On converting the value in cm we get,
⇒r=1cm\Rightarrow r = 1cm
So the correct answer is given by option (A).

Note
In physics, the law of conservation of energy states that energy can neither be created nor destroyed but can only be changed from one form to another. So the kinetic energy of the proton gets converted to potential energy by the law of conservation of energy.