V=Vo(1-e^(-t /RC)?

Vo = 105 , C= 10.54 , R 0.52
(i) Find the value of the voltage after a time 5.
(ii) Find the value of the time when the voltage is half of its eventual value.

Answers:
My guess is that it's 10.54 uF. A 10 farad capacitor that could take 105 V would truly be a sight to see!
This equation describes the decay of voltage across a capacitor connected in parallel with a resistor. It is easily solved with a scientific calculator, and if this sort of problem is likely to be frequent in your life, you should get one: they're very inexpensive nowadays. I use an H-P 42S.
The first answer is correct. However, the values for voltage and resistance are unlikely and that for the capacitor is ridiculous. I think you've missed out some factors of ten which will change the answers - a lot!
e.g. it could be C=10 pF, R=0.52 k ohm
i. V=Vo(1-e^(-t /RC)
Vo = 105 , C= 10.54 , R 0.52
t = 5 secs
Plug in all the given values then V = 62.84 Volts


ii. ln[V/Vo] = ln{e^[-t/RC]}

ln[1/2] = -t/RC
solve for t

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