Question

• A student finds two ancient meters with numbered scales but no unit markings. He thinks that one is an ammeter and one is a voltmeter but doesn't know which is which. He measures the resistance of each, finding that meter A has a resistance of 500 ohms and meter B has a resistance of 0.10 ohms. He can conclude that _______________

A) A= voltmeter B= ammeter

B) A= not useful B= voltmeter

C) A= voltmeter B= not useful

• The plates of a parallel-plate capacitor are 3.28mm apart, and each has an area of 15.7cm^2. Each plate carries a charge of magnitude 0.9nC. The plates are in a vacuum. What is the potential difference between the plates?

A) 212.46 V

B)222.16 V

C) 228.24 V

• The two plates of a capacitor are given charges plusminusQ. The capacitor is then disconnected from the charging device so that the charges on the plates cannot change, and the capacitor is immersed in a tank of oil. What happens to the electric field between the plates?

A) increases

B) decreases

C) becomes zero

Answer 1

1.Since The resistance of voltmeter is greater the ammeter.

so A is voltmeter, B is ammeter.

2.Given, $d=3.28mm, A=15.7cm^2, q=0.9 nc$

Capacitance $C=\dfrac{A\epsilon_o}{d}=\dfrac{15.7\times 10^{-4}\times 8.854\times 10^{-12}}{3.28\times 10^{-3}}=42.30\times 10^{-13} F$

Potential difference $V=\dfrac{q}{C}=\dfrac{0.9\times 10^{-9}}{42.30\times 10^{-13}}=0.0212\times 10^{4}=212volts$

3.When we immersed capacitor in a oil, Then the capacitor is filled with a a dielectric solution of soil, Which has certain dielectric constant value. So Capacitance increases since It is directly proportional to

dilelctric constant.