# Math Help - rates of change

1. ## rates of change

It has been found through the series of experiments that voltage V at the semiconductor diode is related to the charge Q at the diode by the formula V(Q) = V0 (100.434Q/Q0 1), where Q0 and V0 are some constant reference values measured in Coulombs and Volts, correspondingly.

(a) Find the ratio of V/V0 when the charge Q is 10% greater than Q0. Round off your answer to one decimal place.

(b) Which charge (in Coulombs) on the diode is when the voltage is 0 Volts? Present your answer with one decimal place.

(c) Assuming that V0 = 0.001 V, find what is the voltage in the diode when its charge is 10 times greater than the reference charge Q0. Round off your answer to three decimal places.

2. ## Re: rates of change

Originally Posted by jesscarter24
It has been found through the series of experiments that voltage V at the semiconductor diode is related to the charge Q at the diode by the formula V(Q) = V0 (100.434Q/Q0 1), where Q0 and V0 are some constant reference values measured in Coulombs and Volts, correspondingly.

(a) Find the ratio of V/V0 when the charge Q is 10% greater than Q0. Round off your answer to one decimal place.

(b) Which charge (in Coulombs) on the diode is when the voltage is 0 Volts? Present your answer with one decimal place.

(c) Assuming that V0 = 0.001 V, find what is the voltage in the diode when its charge is 10 times greater than the reference charge Q0. Round off your answer to three decimal places.
you've posted this same question a couple of times ... no one is responding because your formula syntax is rubbish.

V(Q) = V0 (100.434Q/Q0 – 1)
... is there an exponent somewhere in this equation?

could the equation actually be ...

$V(Q) = V_0 \left(10^{\frac{0.434Q}{Q_0}} - 1\right)$

???

3. ## Re: rates of change

sorry yes the equation you just wrote is the correct one, wasnt sure how to write in exponents

4. ## Re: rates of change

(a) and (c) are all set up ... get out your calculator.

(b) is very simple if you understand exponents (doesn't even require a calculator)

(a) $\frac{V}{V_0} = 10^{0.434(1.1)} - 1$

(b) if $V = 0$ , then $10^{\frac{0.434Q}{Q_0}} - 1 = 0$

$Q = ?$

(c) $V = 0.001 \left(10^{0.434(10)} - 1\right)$

5. ## Re: rates of change

thankyou,

i had actually figured out a) and c) but just needed clarification.
part b) is the hard part and i can't reaarange the equation to get Q.
i had gotten that far but got stuck there

6. ## Re: rates of change

Originally Posted by jesscarter24
thankyou,

i had actually figured out a) and c) but just needed clarification.
part b) is the hard part and i can't reaarange the equation to get Q.
i had gotten that far but got stuck there
one more time ...

$10^{some \, power} - 1 = 0$

what would "some power" have to be?

THINK!

7. ## Re: rates of change

the power is 0.434Q/Qo , but when i try to get Q on its own, i get Q=0, which i dont think can be right

8. ## Re: rates of change

Originally Posted by jesscarter24
the power is 0.434Q/Qo , but when i try to get Q on its own, i get Q=0, which i dont think can be right
think again ...

lesson one about electricity ... no charge = no voltage

9. ## Re: rates of change

so what would be the next step?

10. ## Re: rates of change

Originally Posted by jesscarter24
so what would be the next step?
what "next step" ?

Q = 0 ... you're done.

11. ## Re: rates of change

so sub in Q=0 and get -1?

12. ## Re: rates of change

Originally Posted by jesscarter24
so sub in Q=0 and get -1?
no, jess ...

the question wants to know what charge $Q$ makes the voltage $V = 0$

$V = V_0 \left(10^{\frac{0.434Q}{Q_0}} - 1\right)$

$0 = V_0 \left(10^{\frac{0.434Q}{Q_0}} - 1\right)$

since $V_0 \ne 0$ ...

$10^{\frac{0.434Q}{Q_0}} - 1 = 0$

$Q = 0$ makes the left side of above equation = 0

in other words, $10^0 - 1 = 0$

13. ## Re: rates of change

oh thankyou got it now! it is just Q=0