Question

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A charged particle moving in the presence of both an electric and magnetic field moved in a closed path (i.e. a path that begins and ends at the same point). If the energy of the particle at the end of the closed path is 2.0J greater than its energy when it began the path, which of the following statements is correct?

a. The magnetic field performed 2.0 J of work on the particle

b. The magnetic field performed -2.0J of work on the particle

c. The electric field is conservative

d. The electric field is due to static charges

e. None of the above

A dielectric material is inserted between the plates of an isolated charged capacitor. Which of the following is true?.

a. The potential energy of the capacitor increases

b. The charge of the capacitor decreases

c. The capacitance of the capacitor decreases

d. The voltage of the capacitor increases

e. None of the above

Consider an infinite line of positive charge that has a varying charge density. Very close to the line of charge, in which direction does the electric field approximately point?

a. Directly toward the line charge

b. Directly away from the line charge

c. Parallel to the line charge, pointing in the direction of increasing charge density

d. Parallel to the line of charge, pointing in the direction of decreasing charge density

e. The field is zero

While discharging a capacitor through a resistor, how many of the following functions are exponentially decreasing? I. the current through the circuit II. The energy stored in the capacitor III. The voltage across the resistor IV. The power dissipated in the resistor V. The electric field within the capacitor VI The charge on the capacitor

a. two

b. three

c. four

d. five

e. six

Answer 1

PL 02 Ans option'e (None of the above) Reason 0 Magnetic force always acts perpendi- cular to velocity of charge particle So, work done by magnetic force is zero ③ Here Electric field is non-conservative in nature, because work done by conservative Electric field is zero in close loop. In this case Electric field is doing work = 205. option 'e (None of the above), Reason → When dielectric is inserted then Capacitance increased, so g=cv so, increase in Capacitance will decrease Voltage and hemce potential energy And option t' (directly away from the Ans - or, uza live charge) -t é RC so, = 20 e E= v while discharging q=% V - q - the [gan - To -t/RC (Electric a cd held] + 9 e RC Energy RC Voltage Across Resisistor a IXR = q et hoc So, All options are correct. I=-da -t = 92 22-2t =% e Re at 2c 2c