LMS Electric Field Quiz
DISCLAIMER: Although every reasonable effort has been taken to ensure the accuracy of the answers provided herein, at the time of posting, I do
not accept liability for any damages, which might arise as a result of the reading, sharing and using of the content provided herein. Users are to
make use of the content provided responsibly, at their own discretion.
1
Point charges, Q1 = +52 nC and Q2 = -91 nC, are
placed as shown. In the figure shown, an external
force transports an electron from point A to
point B. The work done by the external force is
_______.
-12 2 -1
-2
(ε0 = 8.85 x 10 C N m )
Ans +860 eV
Work done is the change in total U of the 3 charges. Since distance between Q 1 & Q2 remains
unchanged, their U also remains the same. So just calculate (ΔU of Q1 & e-) + (ΔU of Q2 & e-).
(Note: ΔU = Ufinal – Uinitial )
2
A point charge Q = -12 μC, and two other charges, q1 and q2, are
placed as shown. The electric force components on charge Q are
Fx = +0.005 N and Fy = -0.003 N. The charge q1, in nC, is _______.
-12
(ε0 = 8.85 x 10
2
-1
-2
C N m )
Ans +600
The force exerted by q1 on Q is the x-component of the electric force on Q. Its positive value (+0.005 N)
indicates that it is acting towards q1, which means that the force is attractive, thereby implying that q1
is a positive charge. Substitute the values into Coulomb’s Law directly to solve for q 1.
3
Ans
When two point charges are a distance d apart, the electric force that each one feels from the other
has magnitude F. In order to make this force twice as strong, the distance would have to be changed
to ________.
d
√2
By Coulomb’s Law, F ∝
1
. When F x 2, d x
d2
1
√.
2
4
2 large conducting parallel plates A and B are separated by 2.4 m. A
uniform field of 1500 V/m in the positive x-direction, is produced by
charges on the plates. The centre plane at x = 0.0 m is an equipotential
surface on which V = 0. An electron is projected from x = 0.0 m with an
initial kinetic energy K = 300 eV in the positive x-direction as shown. The
electric potential difference VA - VB is ______.
Ans +3600 V
Since the direction of an electric field is always towards the lower potential, VA > VB ⇒ VA – VB > 0
For parallel plates, V = E x r
5
A conducting sphere of radius R carries an excess positive charge and is very far from any other charges.
Which one of the following graphs best illustrates the potential (relative to infinity) produced by this
sphere as a function of the distance r from the centre of the sphere?
A
B
D
C
E
Ans B
For a solid conductor, the electric potential inside it would be equal to that on its surface.
6
In the figure shown, 4 charges are placed at the corners of a
square of side 2 cm. What is the electrostatic energy of this
system?
-12
(ε0 = 8.85 x 10
2
-1
-2
C N m )
Ans -4.7 J
Each pair of charges contribute to the total electrical potential energy of the system. Since there are 4
4
point charges, there are C2 = 6 pairs altogether – 4 of them along the sides and 2 of them diagonally
across. Add up the potential energies of all 6 pairs to find the total potential energy.
7
Which of the following statements about electric field lines due to a charge arrangement are correct?
(There may be more than one correct choice.)
(a) If the electric force on an electron points upward at a point, the direction of the electric field is
downward at that point.
(b) They point away from negative charges.
(c) Charges must move along these field lines.
(d) They can never cross.
(e) The more closely the lines are spaced, the stronger the electric field.
Ans a, d, e
Charges do not need to move along the field lines. The lines are just a visual representation of the
magnitude and direction of electric forces.
8
The point charges -3.00 µC, +8.00 µC, and +10.0 µC on a large horizontal table are temporarily held at
the vertices of an equilateral triangle that is 5.00 cm on each side. After they are released and are free
to move with no friction, the maximum total amount of kinetic energy they could gain is ______.
Ans 4.68 J
The maximum amount of kinetic energy that the point charges can gain is equal to the total electric
potential energy of the system. Each pair of charges contributes to this total energy. Since there are 3
3
charges, there are 3 unique pairs ( C2 = 3). The sum of potential energies of each pair is the total
potential energy of the system.
9
The electric field lines arising from 2 charges Q1 and Q2 are
shown in the figure. From this drawing we can see that:
(a) none of these is true.
(b) |Q1| > |Q2|
(c) the electric field could be zero at P1.
(d) both Q1 and Q2 have the same sign.
(e) the electric field could be zero at P2.
Ans a
Since the electric field line diagram is symmetrical about P1 , the number of field lines that ‘leave’ a
charge is equal to the number of lines that ‘reach’ the other charge, thereby implying that both charges
are of equal magnitude and opposite charge.
A point whereby there is ‘zero electric field’ can exist only if the 2 charges are like charges, causing
repulsive forces to create a null point, which would usually be in the centre.
10
A 50-g insulating sphere carries a charge Q = -60 μC and is suspended by a silk thread from a fixed point.
An external electric field which is uniform and vertical is applied. The applied electric field has a
magnitude of 3000 N/C and is directed downward. The tension in the thread is _________.
Ans 0.3 N
Tension = weight + electric force
= mg + qE
11
A point charge Q is located a short distance from a point charge 3Q, and no other charges are
present. If the electrical force on Q is F, the electrical force on 3Q is_______.
Ans F
rd
Newton’s 3 Law
12
Two large conducting parallel plates A and B are separated by 2.4 m. A
uniform field of 1500 V/m in the positive x-direction, is produced by
charges on the plates. The centre plane at x = 0.0 m is an equipotential
surface on which V = 0 V. An electron is projected from x = 0.0 m with an
initial kinetic energy K = 300 eV in the positive x-direction as shown. Find
the initial velocity of the electron at x= 0.0 m.
(Mass of an electron = 9.11 x 10
-31
kg; 1 eV = 1.6 x 10
-19
J)
Ans 1.0 x 107 m/s
1
Ek = 2 mv 2
13
Four charges of equal magnitudes but opposite signs are arranged at the corners of a square, as
shown here. In which arrangement is the magnitude of the electric field at point P a maximum?
a
b
c
d
Ans c
In a & b, point P would be a null point – 0 net electric force – due to repulsion. In c, the +q at the top
left corner repels the +q at the top right corner, causing the density of field lines at P to be higher.
14
5
A proton with a speed of 3.0 x 10 m/s falls through a potential difference V and thereby increases its
5
speed to 9.0 x 10 m/s. Through what potential difference did the proton fall?
(Charge of a proton = 1.6 x 10
-19
C; Mass of a proton = 1.67 x 10
-27
kg)
Ans 3800 V
Potential difference can be considered as energy per unit charge (1 V = 1 J/C). Hence, it is equal to the
kinetic energy gained by the proton divided by the charge of the proton.
15
A proton is projected toward a fixed nucleus of charge +Ze with velocity v0. Initially the 2 particles are
1
very far apart. When the proton is a distance R from the nucleus its velocity has decreased to 2v0 How
1
far from the nucleus will the proton be when its velocity has dropped to 4v0?
Ans 4
R
5
1
Ek = 2 mv 2
Ek ∝ v 2
U=
U∝
Qq
4πε0 r
𝟏
𝒓
From ∞ to R
1
When v0 x 2,
Ek x
1
4
3
U increases from 0 to Ek
(Law of Conservation of Energy)
4
From R to new distance
1
1
1
When (v0 x 2) x 2 = v0 ,
1
1
1
4
4
16
(Ek x ) x =
4
Ek
U increases from
3
4
Ek =
𝟏𝟐
𝟏𝟔
Ek to
Ux
∴Rx
Prepared by:
Muhammed Anas s/o Tariq, 1SB1
5
4
𝟒
𝟓
𝟏𝟓
𝟏𝟔
Ek