A charging capacitor Note: I may ask you to also draw arrows that represent the current at each of these times (see e.g. Ch 19, P34 in Matter and Interactions). •P36 When a particular capacitor, which is initially uncharged, is connected to a battery and a small light bulb, the light bulb is initially bright but gradually gets dimmer, and after 45 s it goes out. The diagrams in Figure 19.71 show the electric field in the circuit and the surface charge distribution on the wires at three different times (0.01 s, 2 s, and 240 s) after the connection to the bulb is made. The diagrams in Figure 19.71 show the pattern of electric field in the wires and charge on the surface of the wires at three different times (0.01 s, 8 s, and 240 s) after the connection to the battery is made. Which of the diagrams best represents the state of the circuit at each time specified? (a) 0.01 s after the connection is made, (b) 8 s after the connection is made, (c) 240 s after the connection is made (A) (B) Enet Ēnet = 0 Figure 19.71 (C) Enet 00
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Solution:-
Answers:-
(A)At t = 0.01 s. ( Situation is described in part (C)
Explanation:-
0.01s is just after switch is closed and current is allowed to pass through the bulb and the capacitor. At this stage bulb will be brightest among all three.
The current will start flowing from (+ve) side of battery to the negative side. The electric field direction will be towards the negative terminal of the battery i.e in the direction of current flow.
As the switch is just closed there will be no charge on the capacitor plate.
(B) At t = 8 s situation is represented by (A)
Explanation:-
t = 8 s is less than 45 s, thus at this stage of time the bulb will glow but dimmer than in case (C).
The electric field will be from (+) ve terminal of battery to the negative terminal of battery. The direction is always towards the (-) terminal.
Current flow direction will be same as the direction of electric field.
At this stage capacitor will be charged. The plate connected with + ve terminal will be positively charged and the plate connected with -ve charge will be negatively charged.
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