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Question

2.The electrostatic potential inside a charged spherical ball is given by phi=ar2+b where r is the distance from the centre ,a and b are constants.Then ,the charge density inside the ball is

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Solution

Step 1

Given,

=ar2+b Where a and b are cont.

r = distance from center.

Step 2

we can solve this by using gauss's law = E4πr2=qin0

Step 3

Solve,

given, =ar2+b

Electrical filed E = ddr=2ar

now putting the value of E on gauss's law

E4πr2=qin0

(2ar)4πr2=qin0

8π0ar3=qin

dqindr=24π0ar2

now,

V=43πr3

dVdr=4πr3

Changingdensityη=dqindrxdrdV

=24π0ar24πr2=60a

Hence, the charge inside the ball is

60a



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