Question

A capacitor of capacitance (2.0±0.1)uF is charged to a voltage V = (20±0.5) V .Calculate the charge Q with error limits. plz plz answer fast.

Answer 1

HI friend here is your answer

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We know, Q=CV

First Addition Q=2.1×10^(-6)×20.5=43.05 micro C

Now, Subtraction Q=1.9×10^(-6)×19.5=37.05 micro C

So, Q=43.05uC to 37.05uC

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Answer 2

Concept introduction:

The relationship of the quantity of electric charge held on a wire to an energy barrier is known as capacitance. Self inductance and mutual capacitance are two concepts that are connected to one another.

Given:

Here it is given that A capacitor of capacitance  (2.0±0.1)uF

voltage V = (20±0.5) V.

To find:

We have to find the charge $Q$.

Solution:

According to the question, We know, $Q=CV$

First Addition $Q=2.1*10^{-6}*20.5=43.05 micro C$

Now, Subtraction $Q=1.9*10^{-6}*19.5=37.05 micro C$

So, $Q=43.05uC$ to $37.05uC$.

Final answer:

Hence, the final answer of the question is->

$Q=43.05uC$ to $37.05uC$.

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