Electromagnetism - Model Answers
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FlashcardsExplain how an insulating object can become charged.
Electrons can be transferred from one object to another.
Via friction.
The object gaining electrons becomes negatively charged.
Describe what is meant by an electric field.
A region where a charged particle experiences a force.
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Explain how an insulating object can become charged.
Electrons can be transferred from one object to another.
Via friction.
The object gaining electrons becomes negatively charged.
Describe what is meant by an electric field.
A region where a charged particle experiences a force.
Define electric field strength.
The force per unit positive charge.
State the equation for electric field strength.
E = F/qsource
Where E is the electric field stregth.
F is the electrostatic force.
q is the charge of the charge placed within the field.
State whether the following are vectors or scalars.
Coulomb force
Electric field strength
Coulomb force - Vector
Electric field strength - Vector
Describe the direction of electric field lines.
Electric field lines point in the direction that a positive point charge would experience a force. (*from +ve to -ve).
Draw the electric field around a positive point charge.
Radial field.
Electric field strength decreases with distance from charge (field lnes become further apart).
Field lines meet charge at right angles to its surface.
Arrows point in the direction that a positive point charge would experience a force (away from +).
Draw the electric field around a negative point charge.
Radial field.
Electric field strength decreases with distance from charge (field lnes become further apart).
Field lines meet charge at right angles to its surface.
Arrows point in the direction that a positive point charge would experience a force (towards -).
Show how the electric field vector at A has been constructed.
Step 1 - Draw the electric field vectors sue to the positive and negative charge (E+ and E- respectively).
*Note the vectors are the same length because point A is the same distance from each charge.
Step 2 - Draw the electric field vectors due to the positive and negative charges (E+ and E- respectively).
Show how the electric field at X has been constructed.
Step 1 - Identify the electric field vectors due to charge 1 and charge 2.
*Note that E1 is shorter in length than E2, as the point we are considering is further from charge 1 than charge 2.
Step 2: Vector diagram.
Draw the electric field lines between a positive point charge and nagative place.
Charges coming in all directions.
Define unifrom electric field.
A field in which the electric field strength is constant everywhere.
Draw the electric field between two infinite parallel plates.
Key points to note:
Arrows from + to - (as this is the direction of a +charge would experience.
Field lines equidistant (as the electric field strength is presented by the clossness of field lines, and this is constant).
Field lines meet plates perpendicular to the plate.
State the equation for the electric field in a uniform field.
E = V/d.
Where E was the electric field strength.
V is the potential difference across two plates.,
d is the seperation between the two parallel field.
State Coulomb’s Law.
F = kq1q2/ r2
Where F is the electrostatic force
k is Coulomb’s constant = 8.99 × 109 Nm2C-2
r is the sepeartion between the centres of the charges
Derive the base unit for k.
k = Fr2 / q1q2
[k] = [F] [r]2 / [Q]2 = Nm2C-2
Base units are: m, s, A, kg
So N = kgms-2 and C= As
[k] = kgms-2m2A-2s-2
= kgm3s-4A-2
Describe the relationship between force and separation between charges.
The force is inversely proportional to the square of the separation (inverse square law).
Two identical table tennis balls, M and N, are attached to non-conducting threads and suspended from a point P. The balls are each given the same positive charge and they hang as shown in the diagram. The mass of each ball is 2.7g.
a) Draw a force arrow diagram for charge M.
b) Calculate the charge on each of the charges.
a) Tension in the string, repulsive electrostatic force, weight (mg)
b)
Resolving forces vertically:
Tvertical = mg
T cos35 = 2.7 ×10-3 × 9.81 = 0.026487
T = 0.0323 N
Resolving forces horizontally:
Thorizontal = Felectrostatic = kq1q2 / r2
As each charge has the same charge, q1 = q2 and:
Thorizontal = Felectrostatic = kq2 / r2
0.0323 sin35 = 8.99 × 109 x q2 / (20.6 × 10-2)2
Q = 2.957 × 10-7 C = 3.0 × 10-7 C
Derive the expression for the electric field strength in a radial field.
F = kq1q2 / r2
E = F/q (from the definition of field strength)
E = kq1q2 / r2q
E = kq / r2
Where k is Coulomb’s constant, q is the charge creating the field, r is the distance from the centre of the charge q.
Prove the relationship between electric field strength and separation, as shown in this graph, obeys the inverse square law.
If E and r follows the inverse square law, then Er2 should be a constant as E = constant / r2
E1r12 = 11 ×1011 x (3.6 ×10-11)2 = 1.43 ×10-9 NC-1m2
E2r22 = 4 ×1011 x (6.0 ×10-11)2 = 1.44 ×10-9 NC-1m2
E3r32 = 2 ×1011 x (8.4 ×10-11)2 = 1.41 ×10-9 NC-1m2
Er2 is a constant to 2sf, hence E and r follow the inverse square law.
Determine the resultant electric field strength at position X.
Let the 60μC charge be Q1 and the -80μC charge be Q2.
E1 = kQ1/ r12 = 8.99 ×109 × 60 ×10-6 / (12 ×10-3)2 = 3.75 ×109 NC-1
E2 = kQ2/ r22 = 8.99 ×109 × 80 ×10-6 / (9 ×10-3)2 = 8.88 ×109 NC-1 (ignore sign)
Magnitude of resultant (use Pythagoras):
Etotal = √(3.75 ×109)2 + (8.88 ×109)2 = 9.64 ×109 NC-1
Direction (use trigonometry):
θ = tan-1 (8.88 ×109 / 3.75 ×109 )
= 67.1 degrees below horizontal
Define electric potential.
The work done per unit positive charge, in bringing a charge from infinity to a point within the field.
Explain why the electric potential is positive for positive charges and negative for negative charges.
In bringing a positive test charge from infinity to a point within the field around a positive charge, work has to be done against the field.
This is because the test charge would experience a repulsive electrostatic force in the opposite direction to its displacement.
So the test charge would gain electrical potential energy as it moves closer to the field creating positive charge (transferring it from kinetic energy).
Whereas when a positive test charge is brought closer to a negative field creating charge, the test charge would experience an attractive electrostatic force towards the negtive field creator.
Therefore work would be done by the field.
The charge’s electrical potential energy would decrease (it would be transferred into kinetic energy) as it gets closer to the negative charge.
Sketch a graph of potential against separation from
a) a positive charge
b) a negative charge
Inversely proportional graph for positive region - Positive charge
Reflected in x axis - Negative charge
Describe the relationship between electric potential, V and seperation, r.
Potential and separation are inversely proportional.
Determine the potential at position X due to charges Q1 and Q2.
Vtotal = V1 + V2
Let the 60μC charge be Q1 and the -80μC charge be Q2.
V1 = kQ1/ r1 = 8.99 ×109 × 60 ×10-6 / 50 ×10-3 = 10,788,000 V
V2 = kQ2/ r2 = - 8.99 ×109 × 80 ×10-6 / 40 ×10-3 = - 17,980,000 V (KEEP SIGN)
Vtotal = V1 + V2 = -7,192,000 = 7.192 ×106 V
State the equation showing the relationship between electric potential and field strength.
E = -∆V / ∆r
E is the electric field strength
∆V is the change in potential for a charge, in distance ∆r
David Blaine wore a metal suit and was placed in a room with a charged sphere that cuased ionisation at 1m. A scientist claims that David Blaine is completely safe inside the metal cage. Comment on this suggestion. [4 marks].
There is no potential difference across the cage or his body.
Electric field strength inside the cage is zero (it is a Faraday Cage).
This is because there is no potential gradient.
Current conducts only on the surface of the cage - not inside.
Determine the electric field strength between X1 and X2.
E = ∆V/ ∆r = (600 - 400)/ (10 - 5) x10-2 = 4000 Vm-1
Explain how to determine the electric field strength from a graph of potential against separation.
Electric field strength is the nagtive gradient of a potential, distance graph.
If the potential, distance graph is a curve, then the tangent at a point should be drawn and the gradient of this should be found.