Question

• Applications of Gauss’s Law

Q. A hollow metal sphere of radius R is uniformly
charged. The electric field due to the sphere at a
distance r from the centre

(a) decreases as r increases for r < R and for r > R

(b) increases as r increases for r < R and for r > R

(c) zero as r increases for r < R, decreases as r
increases for r > R

(d) zero as r increases for r < R, increases as r
increases for r > R

$(NEET \: 2019)$
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Answer 1

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Q. A hollow metal sphere of radius R is uniformly

charged. The electric field due to the sphere at a

distance r from the centre

(a) decreases as r increases for r < R and for r > R

(b) increases as r increases for r < R and for r > R

(c) zero as r increases for r < R, decreases as r

increases for r > R

(d) zero as r increases for r < R, increases as r

increases for r > R

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The correct option is (2) Zero as r increases for r < R, decreases as r increases for r > R. Explanation: Charge Q will be distributed over the surface of hollow metal sphere.

HOPE IT HELPS YOU SIR :)

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Answer 2

Gauss's law:

It states that the flux via a closed surface is directly proportional to charge it encloses.

OR we can simply say that..

• Whenever any charge is enclosed by closed surface , Electric field will be present within it.
• No charge, No Electric field.

Solution:

Question clearly says METAL Hollow sphere.

• means it's conductor.
• and by property,all charge will uniformly get distributed throughout the surface of sphere.
• Charge inside = 0

Now, consider a spherical closed surface of radius r < R, (attached image)

• charge enclosed within it is zero.
• no charge, No EF
• EF will be zero throughout the sphere as r <R.

Consider a spherical closed surfaces outside the sphere at increasing distance r > R,

• sphere will be acting as point charge source.
• Field will decrease with increasing distance.

★ so, (c)zero as r increases for r < R, decreases as r increases for r > R is correct.

By above we can draw graph(attached) and can place the general formulas for hollow sphere:

• $\rm E_{in}= 0$
• $\rm E_R = \dfrac{Kq}{R^2}$
• $\rm E_{out} = \dfrac{Kq}{r^2}$

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