JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 20)
A positive point charge is released from rest at
a distance r0 from a positive line charge with
uniform density. The speed (v) of the point
charge, as a function of instantaneous distance
r from line charge, is proportional to :-
_8th_April_Evening_Slot_en_20_1.png)
$$v \propto \left( {{r \over {{r_0}}}} \right)$$
$$v \propto \ln \left( {{r \over {{r_0}}}} \right)$$
$$v \propto {e^{ + r/{r_0}}}$$
$$v \propto \sqrt {\ln \left( {{r \over {{r_0}}}} \right)} $$
Explanation
$${1 \over 2}m{V^2} = - q\left( {{V_f} - {V_i}} \right)$$
$$E = {\lambda \over {2\pi {\varepsilon _0}r}}$$
$$\Delta V = {\lambda \over {2\pi {\varepsilon _0}}}\ln \left( {{{{r_0}} \over r}} \right)$$
$${1 \over 2}m{v^2} = {{ - q\lambda } \over {2\pi {\varepsilon _0}}}\ln \left( {{{{r_0}} \over r}} \right)$$
$$v \propto \sqrt {\ln \left( {{r \over {{r_0}}}} \right)} $$
$$E = {\lambda \over {2\pi {\varepsilon _0}r}}$$
$$\Delta V = {\lambda \over {2\pi {\varepsilon _0}}}\ln \left( {{{{r_0}} \over r}} \right)$$
$${1 \over 2}m{v^2} = {{ - q\lambda } \over {2\pi {\varepsilon _0}}}\ln \left( {{{{r_0}} \over r}} \right)$$
$$v \propto \sqrt {\ln \left( {{r \over {{r_0}}}} \right)} $$
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