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Chapter 19: Q76P (page 802)

The deflection plates in an oscilloscope are 10cm by 2cm with a gap distance of 1mm. A 100V potential difference is suddenly applied to the initially uncharged plates through a 1000鈥�惟resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95V?

Short Answer

Expert verified

t=5.3脳10-8s

Step by step solution

01

Given data

Plate area A=10鈥�cm脳2鈥�cm

Gap s=1鈥�mm

Potential difference V=95鈥�V

Max. Potential difference Vmax=100鈥�V

Resistance R=1000鈥�惟

02

Formula used

V=Vmax(1-exp-tRC)

Where,Vmaxis max. Potential difference reaches

t Is time require

R is resistance andC is the capacitance

03

Calculation for time required

C=蔚0As=8.85脳10-12脳2脳10-31脳10-3=1.77脳10-11鈥�F

V=Vmax1-exp-tRC95=1001-exp-t1000脳1.77脳10-110.95=1-exp-t1.77脳10-8exp-t1.77脳10-8=0.05

t=5.3脳10-8s

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Most popular questions from this chapter

Using thick connecting wires that are very good conductors, a Nichrome wire (鈥渨ire 1鈥�) of length L1 and cross-sectional area A1 is connected in series with a battery and an ammeter (this is circuit 1). The reading on the ammeter is I1. Now the Nichrome wire is removed and replaced with a different wire (鈥渨ire 2鈥�), which is 2.5 times as long and has 5.5 times the cross-sectional area of the original wire (this is circuit 2). In the following question, a subscript 1 refers to circuit 1, and a subscript 2 refers to circuit 2. It will be helpful to write out your solutions to the following questions algebraically before doing numerical calculations. (Hint: Think about what is the same in these two circuits.)(a) What is the value of I2/ I1, the ratio of the conventional currents in the two circuits? (b) What is the value of R2/ R1, the ratio of the resistances of the wires? (c) What is the value of E2/ E1, the ratio of the electric fields inside the wires in the steady states?

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