A radioactive source in the form of a metal sphere of radius 10^-2 m, emits beta particles at the rate of 5 x 10¹⁰partiCles per sec. The source is electrically insulated. How long will it take for it's potential to be raised by 2 volts, assuming 40% of the emitted beta particles escape the source

from where do u got this question.?

Ans is (approx.) 2.778 * 10^-4 sec
i m not sure pl tell me..i will tell u the method.

i was checking out tricky questions on google on electrostatics and i found it..it dint show the solution how did you get the answer?

is the answer 7x10^-4 seconds?

ok i was wrong think ans is t = 6.944 * 10^-4 sec

i dont know the answer dude..please the the procedure of solving it

we know tat E = - dV / dr
then E = K q / r²
Then
(K q / r²) . dr = -dV
now integrate on both the side
Kq / r = V
Kq = Vr
​we have given V = 2V
​r = 10^-2m
then Kq = 2 * ​10^-2m -------(a)
also we know that q = ne
then  
dq/dt = e * dn/dt
we have given that beta particle is emited that means electrons are emited at the rate of 5 * 10¹⁰ particles per second.
then dn / dt = 5 * 10¹⁰
dq / dt = 5 * 10¹⁰ * e
dq = 5 * 10^{10 *} e  .dt
integrate on both the sides
q = 5 * 10¹⁰ * e *t
but given that 40 % get lost then q becomes 
(40 / 100) * 5 * 10¹⁰ * e * t = 2 * 10¹⁰ * e * t   -----(b)
From eq (a) and eq(b)...... Kq = 2 * ​10^-2
K * ​2 * 10¹⁰ * e * t = 2 * ​10^-2
since K = 9 * 10⁹ and e = 1.6 * 10-19
t = 6.944 * 10^-4 seconds

we know tat E = - dV / dr
then E = K q / r²
Then
(K q / r²) . dr = -dV
now integrate on both the side
Kq / r = V
Kq = Vr
​we have given V = 2V
​r = 10^-2m
then Kq = 2 * ​10^-2m -------(a)
also we know that q = ne
then  
dq/dt = e * dn/dt
we have given that beta particle is emited that means electrons are emited at the rate of 5 * 10¹⁰ particles per second.
then dn / dt = 5 * 10¹⁰
dq / dt = 5 * 10¹⁰ * e
dq = 5 * 10^{10 *} e  .dt
integrate on both the sides
q = 5 * 10¹⁰ * e *t
but given that 40 % get lost then q becomes 
(40 / 100) * 5 * 10¹⁰ * e * t = 2 * 10¹⁰ * e * t  
then remaining electrons = (5-2) * 10¹⁰ * e * t =  3 * 10¹⁰ * e * t   ----(b)
From eq (a) and eq(b)...... Kq = 2 * ​10^-2
K * ​3 * 10¹⁰ * e * t = 2 * ​10^-2
since K = 9 * 10⁹ and e = 1.6 * 10-19
t = 4.62 * 10^-4 seconds

thankyou!!!

the ans is 4.62 * 10^-4 seconds 
sorry i forgot to substract the left electrons..