chapter 22 - electric fields

Question 1

electric field

Answer 1

region of space where a charged object experiences a force due to another charged object

Question 2

observing electric fields

Answer 2

• metal foil between two charged plates
• charged balloons suspended from a point by string
• gold leaf electroscope

Question 3

electric field strength

Answer 3

the force exerted per unit charge on a small positive test charge placed at that point
is a property of a field and not the particular charge in it
E = F/q
unit = N?C
vector - points the direction a positive charge would feel a force
due to coulombs law E (at q) = Qk/r²

Question 4

electric field lines

Answer 4

show electric field strength and direction of the field
- +>-
- perp to surface of a conductor
- closer = stronger
- equally spaced and parallel = uniform field
show direction a + charge would go

Question 5

electric field patterns

Answer 5

+ point charge > radiates out all directions (- opposite)
+ charge and - > goes from + to -
+ and + > radiates out from both - null point in the middle which has no force due to electric field

Question 6

force on a charge due to electric field

Answer 6

F = Eq

Question 7

coulombs law

Answer 7

F ∝ Qq
F ∝ 1/r²
the force between two point charges (q and Q) r apart in a vacuum = kQq/r²
where k = 9x10⁹ Nm²/C² = 1/4𝞹ε
where ε = permittivity of free space

Question 8

coulombs law spheres of charge

Answer 8

if a sphere of charge assume all charge is at the centre and use that assumption to find the distance from one centre to another

Question 9

testing coulombs law

Answer 9

set up a sphere attached to a rod and a sphere suspended from thread of length 1m
measure distance r from the balls at max repulsion and d (r/2)
calculate F as W = Tcosθ
F = Tsinθ
so F = Wtanθ - tanθ = d/1 = d
plot F against 1/r^2
grad = kQq

Question 10

validity of assume charge is at the centre of a sphere

Answer 10

charges will redistribute due to repulsion
like charges will be repulsed to opposite sides so in reality the distance (r) will be larger

Question 11

energy / work done

Answer 11

E = V/d

Question 12

electron gun

Question 13

velocity of electrons accelerated by an electron gun

Answer 13

1/2mv² = qV
v = √2qV/m

Question 14

oscilloscopes

Answer 14

• electron gun produces a beam of electrons
• there are y plates (horizontal) and x plates (vertical)
• pd across the y plates deflects the trace vertically
• pd across x plates deflects the trace horizontally
• light produced on screen by electron beam

Question 15

electric potential energy

Answer 15

work done to bring q from infinity to a defined point (Q)
(work done to overcome the repulsion or attraction between charged points)
energy is transferred to the moving charge (its electric potential increases)
EPE (of q in the field of Q) = Fd = kQq/r

Question 16

EPE when charges are the same

Answer 16

EPE increases (positive work is done bringing a test charge from infinity to a point of positive charge)

Question 17

EPE when charges are opposite

Answer 17

EPE decreases (negative work is done bringing a test charge from infinity to a point of negative charge)

Question 18

electric potential

Answer 18

work done per unit charge in brining a positive test charge from infinity to a point
electric potential energy per unit charge
V = EPE/Q
= kQ/r

Question 19

EPE/ V/ E / F
scalar or vector

Answer 19

EPE / V - scalar - can add up individual potentials to find total potentials of multiple charged
E/F - vector - must consider direction when adding individuals to find resultant

Question 20

equipotential lines

Answer 20

surfaces of constant potential - a test charge moving along an equipotential has constant EPE
represented by dotted lines
no work is done on the charge as it moves along an equipotential as the force is at right angles to the equipotential
right angles to field lines

Question 21

all equations

Answer 21

F (2 charges) = kQq/r²
F (1 charge) = EQ
E (field strength) = V/d
W = QV
E (radial field field strength) = kQ/r²
V = kQ/r = Er