Question

How to solve these two questions? ​

Answer 1

1) See attached figure

The relationship between charge and current is:

$i = \frac{Q}{t}$

where

i is the current

Q is the charge

t is the time

Therefore, the current is the rate of change of the charge passing through a given point over time.

This means that for a graph of charge over time, the current is just equal to the slope of the graph.

For the graph in this problem:

- Between t = 0 and t = 2 s, the slope is

$\frac{50-0}{2-0}=25 C/s$

therefore the current is

i = 25 A

- Between t = 2 s and t = 6 s, the slope is

$\frac{-50-(50)}{6-2}=-25 C/s$

therefore the current is

i = -25 A

- Between t = 6 s and t = 8 s, the slope is

$\frac{0-(-50)}{8-6}=25 C/s$

therefore the current is

i = 25 A

The figure attached show these values plotted on a graph.

2) $15 \mu C$

The previous equation can be rewritten as

$Q = i t$

This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.

Here we have the current vs time graph, so we gave to find the area under it.

The area of the first triangle is:

$A_1 = \frac{1}{2}(0.001 s)(0.010 A)=5\cdot 10^{-6} C$

While the area of the second square is

$A_2 = (0.002 s - 0.001 s)(0.010 A)=1\cdot 10^{-5}C$

So, the total area (and the total charge) is

$Q=A_1 +A_2 = 5\cdot 10^{-6} + 1\cdot 10^{-5} = 1.5\cdot 10^{-5}C=1.5 \mu C$