Electrostatic Potential and Capacitance Part 1

Given, the peak voltage of the battery, V0 = 20V
The voltage of the battery, V = 2V
Time, t=1μs=1×10−6s
Resistance of the capacitor, R = 10 Ω
As we know that,
V=V0(1−e−t∕RC)
Substituting the values in the above equation, we get
2=20(1−e−t∕RC)
⇒

t

RC

=ln(

10

9

)
⇒ C=

t

R‌ln( 10 9 )

=

10−6

10×ln( 10 9 )

=0.95μF
∴ The capacitance of the capacitor is 0.95 μ F.