Question

A capacitor of capacitance $C=2.0\pm 0.1\mu F$ is charged to a voltage $V=20\pm 0.2V$. What will be the charge $Q$ on the capacitor? Use $Q=CV$

• A. $40\pm 2.4\times {10}^{-6}$ coulomb
• B. $50\pm 2.4\times {10}^{-6}$ coulomb
• C. $60\pm 2.4\times {10}^{-6}$ coulomb
• D. $80\pm 2.4\times {10}^{-6}$ coulomb

Answer 1

Answer: (A)

$Q=CV$

$C=(20\pm 0.1)\times {10}^{-6}$ f

$V=20\pm 0.2$ v

Q First we find the Q of original values

$Q=2\times 20=40\times {10}^{-6}C$

Now Error % in C is

$\frac{0.1}{2}\times 100=5\%$

error % in v is

$\frac{0.2}{20}\times 100=1\%$

Total error $=1+5=6\%$

error voltage $=\frac{6}{100}\times (40\times {10}^{-6})=2.4\times {10}^{-6}$

$\therefore Q=(40\pm 2.4)\times {10}^{-6}C$