A capacitor of capacitance C = 3.0 `pm 0.1 muF` is charged to a voltage of V = `18 pm 0.4` Volt. Calculate the charge Q[Use Q = CV].

Explanation

`(C +DeltaC) = (3.0 pm 0.1) muF`
`(V+DeltaV) = (18pm0.4)V`
Q = CV
`Q = 3.0 xx 10^(-6) xx 18 = 54 xx 10^(-6)` coulomb
Error in `V = (DeltaV)/(V) xx 100 = (0.4)/(18) xx 100 = 2.2%`
Error in Q = Error in C + Error in V = 3.3% + 2.2% = 5.5%
`:.` Charge Q = `(54 xx 10^(6) pm 5.5%)` coulomb