This page contains the AQA AS Physics P22 Electrical Fields Questions and kerboodle answers for revision and understanding.This page also contains the link to the notes and video for the revision of this topic.
Banner 1
22.1 Field patterns AQA A2 Physics P22 Electrical Fields Kerboodle Answers: Page No. 361
1.Ans-a) An insulated metal can is given a positive charge by touching it with a positively charged rod. Metals have de-localised electrons which are able to move to the positively charged rod to cancel the charge on the rod. This leaves a fewer electrons back on the metal can and as a result it acquires a positive charge. bThe electrons are mobile and so when the negatively charged sphere is connected to the ground via a wire these mobile electrons will move to the ground and the sphere discharges. 2.Ans- a) i 75nA ii 1.9 x 1011 3Ans-a) Like charged repel. The earthed conductor has undergone polarization with negative and positive charged distributed evenly. The action of the negatively charged object repels the negative electrons of the free conductor. b The earth acts as a source of positive and negative charge. The charged object is brought into proximity with the conductor which is then earthed. The object’s negative charge will attract positive charges from the earth into the object. Once it is unearthed the positive charges cannot leave the conductor, so when the object is removed the positive charge remains.22.2 Electric field strengthAQA A2 Physics P22 Electrical Fields Kerboodle Answers: Page No. 365
1.a) Calculate the magnitude of the force On if the electric field strength where 01 is placed is 3.5 104 Vm-l. 1.4 x 10-3 N b )4.0×10-4 vm-1 2.a)i Describe the type of charge carried by X. negative ii Calculate the charge carried by the particle. 1.3 x 107 c b.i 7.3 x 10-3 N ii State the direction of the force on the particle when it is at Y. Towards the metal surface 3.a The voltage is adjusted to 4.5kV. Calculate: i the electric field strength between the plates, 9.0 x 104 v m-1 ii the electrostatic force on a droplet in the field that carries a charge Of 8.0 x IO- 7.2 x 10-14 N b.80mm 4.Ans a 2.9 kV b i 5.6 x 10-15N22.3 Electric potential AQA A2 Physics P22 Electrical Fields Kerboodle Answers : Page No. 368
I.Ans.a) the potential energy of the electron at i -soy, -8.0 x 10-18 J ii +450″, +7.2 x 10-17 J b +8.0 x 10-17 J 2 a) if F+3.OpC, -1.8 x 10-3 J b +1.2 x 10-3 J 3.a) i 250 V m-1 ii the force on the droplet. 8.0 x 10-17 N (towards the negative plate) b -8.0 x 10-19 J 4 a Electric potential, the amount of work needed to move a unit charge from a reference point to a specific point against an electric field. SI unit is joule per coulomb i.e Volts b i 3000 V m-122.4 Coulomb’s law AQA A2 Physics P22 Electrical Fields Kerboodle Answers : Page No. 371
1a protonat a distance of 2.5* 10-9m, 3.7 x 10-11 N b a nucleus of a nitrogen atom (charge +?e) at a distance Of 2.s 2.6 x 10-10 N 2 a i the distance d between the two charges, 69 mm ii 3.6×10-6 N b A charge of +4.0nC is added to each charge in part a. Calculate the force between 01 and when they are at separation d. 2.5 x 10-5 N repulsion 3 a 6.1 nC, negative b The two charges are moved 2.5 mm further part. Calculate the force between them in this new position. 2.2x 10-2 N 4 a) Calculate the magnitude, O, Of each charge, and State whether the charges attract or repel each other. 2.7 nC, attract b 6.2 x 10-2 m, repel22.5 Point charges AQA A2 Physics P22 Electrical Fields Kerboodle Answers : Page No. 374
1 a 5.3 x 106 V m-1 b 10mm 2 a i 3.7 x 108 V m-1 ii 5.6 x 10-3 N towards Q1 3a Calculate the resultant electric field strength: i 4.5 x 108 V m-1 towards Q2 ii 2.6 x 108 V m-1 away from Q1 b i Explain whu there is a point along the line between the two charges at which the electri field strength is zero. ii. 11 mm from Q1, 9 mm from Q2 4 a -9.0 x 106 V b i Show that the electric potential is zero at a point between the two charges which is 10 mm from and 20 mm from ii 2.0 x 109 V m-1 directly towards Q2 Banner 2Practice questions: Page No. 376-379
(a) (i) The electric field strength at a point is the force per unit charge…acting on a small positive test charge placed at the point. (ii) E is a vector quantity. (b) (i) (ii) (c (i)- Two lines with arrows at P, one representing
- E8 twice as long as E4.
- Arrowed line R, the correct resultant of the parallelogram, drawn through the common point of E8 and E4.
- The electron loses potential energy as it moves away from Q.
- As it moves away its speed will increase.
- Its de Broglie wavelength will decrease…
- because its momentum will increase.
- (i)
- Diagram drawn to show:
- An arrow, horizontal to the right, representing vH (or 20 m s−1) and an arrow from the same point, vertically downwards, representing vV.
- An arrow from the same point, at 35° to the horizontal, along the diagonal of the parallelogram representing the resultant velocity v.
- (i)
- (i)
- At B, field strength E = 21 N C–1 (or V m–1)
- Starts at (6.4, 84) and is a curve of decreasing gradient …
- passes through (12.8, 21) and does not meet or intersect r axis
- (i)