A thick spherical shell of inner and outer radii $r_{1} a n d r_{2}$ is made of a material of resistivity $ρ$ and dielectric constant K. Initially the shell is electrically neutral and has no charge accumulation anywhere. If at the instant t=0, a point charge $q_{0}$ is some how placed at the centre, then charge on the outer surface of the shell as a function of time t. (ignore any effects due to magnetism of radiation)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Time constant for the charging process is $ρ ε_{0}$

Time constant for the charging process is $ρ K ε_{0}$

charge on the outer surface of the shell as a function of time t is $q_{0} (1 - e^{\frac{- t}{ρ ε_{0}}})$

charge on the outer surface of the shell as a function of time t is $q_{0} (1 - e^{\frac{- t}{K ρ ε_{0}}})$

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$E 4 π r^{2} = \frac{q_{0} - q_{t}}{K ε_{0}} j = \frac{1}{ρ} E j = \frac{1}{ρ} \frac{1}{4 π r^{2}} (\frac{q_{0} - q_{t}}{K ε_{0}}) j 4 π r^{2} = \frac{1}{ρ K ε_{0}} (q_{0} - q_{t}) \frac{d q t}{d t} = \frac{1}{ρ K ε_{0}} (q_{0} - q_{t}) \int \frac{d q t}{q_{0} - q_{t}} = \int \frac{d t}{ρ K ε_{0}} - \ln (\frac{q_{0} - q_{t}}{q_{0}}) = \frac{t}{ρ K ε_{0}} q_{t} = q_{0} (1 - e^{\frac{- t}{ρ K ε_{0}}})$