Physics, 31.03.2020 04:24 alexabessin

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 975 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 75 volts

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Answer from: lia3865

t = 23.9nS

Explanation:

given :

Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

distance d= 1mm=> 0.001

resistor R= 975 ohm

Capacitance can be calculated through the following formula,

C = (ε0 x A )/d

C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

C = 17.7 x 10^-12 (pico 'p' = 10^-12)

C = 17.7pF

the voltage between two plates is related to time, There we use the following formula of the final voltage

Vc = Vx (1-e^-(t/CR))

75 = 100 x (1-e^-(t/CR))

75/100 = (1-e^-(t/CR))

.75 = (1-e^-(t/CR))

.75 -1 = -e^-(t/CR)

-0.25 = -e^-(t/CR) --->(cancelling out the negative sign)

e^-(t/CR) = 0.25

in order to remove the exponent, take logs on both sides

-t/CR = ln (0.25)

t/CR = -ln(0.25)

t = -CR x ln (0.25)

t = -(17.7 x 10^-12 x 975) x (-1.38629)

t = 23.9 x $10^{-9$

t = 23.9ns

Thus, it took 23.9ns for the potential difference between the deflection plates to reach 75 volts

Answer from: davelopez979

t = 24.3ns

Explanation:

The deflection plates make up a parallel plate capacitor fed via 975 ohm resistor from a 100V supply

capacitance C = (e0 x A) / d

C = $\frac{(8.85 X 10^{-12} X (2 X 10^{-2} X 10 X 10^{-2} ) )}{10^{-3} }$

C = $C = 18 X 10^{-12} F$

$Vc = Vx (1-e^{-(\frac{t}{CR} )} )$

$75 = 100 X (1-e^{-(\frac{t}{CR} )} )$

$\frac{75}{100} = (1-e^-{(\frac{t}{CR} )} )$

$0.75 = (1-e^{-(\frac{t}{CR} )} )$

$0.75 - 1 = -e^{-(\frac{t}{CR} )}$

$e^{-(\frac{t}{CR} )} = 0.25$

Taking the logs of both sides

$-(\frac{t}{CR} ) = In(0.25)$

$(\frac{t}{CR} ) = -In(0.25)$

t = =CR x In (0.25)

$t = (18 X 10^{-12} X 975) X (-1.386)$

t = 2.43 x 10∧-9

t = 24.3ns

Answer from: Quest

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

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Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with...