The area of a blot of ink is growing such that after t seconds, its area is given by `A = (3t^(2) +7) cm^(2)`. Calculate the rate of increase of area at t=5 second.

Answer

(i) `4a, (5a, 0)` (ii) `V=(Q)/(4pi in_(0))(1/(3a-x)-2/(3a+x))` for `x le 3a, Q/(4pi in_(0)) (1/(x-3a)-2/(3a+x))` for `x gt 3a`
(iii) `sqrt((Qq)/(8pi in_(0) ma))`

Explanation

`[A] V_(S)=0=(KQ)/sqrt((x-3a)^(2)+y^(2))-(2KQ)/sqrt((x+3a)^(2)+y^(2))`
`rArr (x-5a)^(2)+y^(2)=(4a)^(2) :. C=(5a, 0)` and radius `=4a`
`[B] V_((x))=(KQ)/(x-3a)-(2KQ)/(x+3a)` for `x gt 3a`
`=(KQ)/((3a-x))-(2KQ)/(3a+x)` for `x lt 3a`
[C] A positive charge when released will move from high potential to low potential
COME: `K_(1)+U_(1)=K_(2)+U_(2)`
`0+q[(KQ)/(2a)-(2KQ)/(8a)]=1/2 mv^(2)+0 rArr V=sqrt((Qq)/(8pi in_(0) ma))`