Question

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

Answer 1

Answer:

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 2

Answer:

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