JEE Advanced 2018 Paper 1 · Physics · Electrostatics · hard

A non-conducting solid sphere of radius 0.20 m carries a total charge of 2.0 muC distributed uniformly throughout its volume. A proton is released…

Numerical answer

A non-conducting solid sphere of radius

0.20\ \mathrm{m}

carries a total charge of

2.0\ \mu\mathrm{C}

distributed uniformly throughout its volume. A proton is released from rest at the centre of the sphere. Neglect any resistance to the proton's motion through the sphere and ignore its effect on the charge distribution. Use

1/(4\pi\varepsilon_0)=8.99\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}

e=1.602\times10^{-19}\ \mathrm{C}

and

m_p=1.673\times10^{-27}\ \mathrm{kg}

. Find the speed of the proton, in units of

10^{6}\ \mathrm{m\,s^{-1}}

, when it reaches the surface of the sphere. Give the answer correct to two decimal places.

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Answer: 2.93

Inside a uniformly charged sphere the field is

E=\dfrac{Qr}{4\pi\varepsilon_0R^3}

. The potential difference between the centre and the surface is

\Delta V=\displaystyle\int_0^R E\,dr=\dfrac{Q}{8\pi\varepsilon_0R}

. With

Q=2.0\times10^{-6}\ \mathrm{C}

and

R=0.20\ \mathrm{m}

\Delta V=\dfrac{8.99\times10^{9}\times2.0\times10^{-6}}{2\times0.20}=44.95\ \mathrm{kV}

. The proton moves outward down the potential, so energy conservation gives

\tfrac12m_pv^2=e\Delta V

. Then

v=\sqrt{\dfrac{2\times1.602\times10^{-19}\times44950}{1.673\times10^{-27}}}\approx2.93\times10^{6}\ \mathrm{m\,s^{-1}}

. The speed is well below

c

, so the non-relativistic treatment is justified.

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