JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Electrostatics

There is a uniform spherically symmetric surface charge density at a distance

R_{0}

from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed

V (R (t))

of the distribution as a function of its instantaneous radius

R (t)

is:

(A)(see structure)
(B)(see structure)
(C)(see structure)
(D)(see structure)
✓ Correct

Answer: (D)

Solution

Use energy conservation on the expanding shell. A uniformly charged shell of charge

Q

and radius

r

stores electrostatic energy

U = \frac{k Q ^{2}}{2 r}

. Starting from rest at radius

R_{0}

and reaching radius

R

with speed

v

\frac{k Q ^{2}}{2 R _{0}} = \frac{k Q ^{2}}{2 R} + \frac{1}{2} m v^{2}

, which gives

v = \frac{k Q ^{2}}{m} (\frac{1}{R _{0}} - \frac{1}{R})

. Reading the shape of this curve: at

R = R_{0}

the speed is zero, so the graph begins at the point

(R_{0}, 0)

; as

R

grows the speed rises but ever more slowly, and as

R \to \infty

it levels off at the finite value

\frac{k Q ^{2}}{m R _{0}}

. The matching graph is the one that starts at

R_{0}

on the horizontal axis, increases while bending over, and approaches a horizontal asymptote.

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Electrostatics

Solution

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