Question

A capacitor of capacitance C = 2.0 ± 0.1 μF is charged to a voltage V= 20 ± 0.2 V. What will be the charge Q on the capacitor? Use Q= CV

Answer 1

Answer:

Q=CV

C=(20±0.1)×10

−6

f

V=20±0.2 v

Q First we find the Q of original values

Q=2×20=40×10

−6

C

Now Error % in C is

2

0.1

×100=5%

error % in v is

20

0.2

×100=1%

Total error =1+5=6%

error voltage =

100

6

×(40×10

−6

)=2.4×10

−6

∴Q=(40±2.4)×10

−6

C