Practice ES2: Electrostatic Potential and Electric
Potential with Point Charges Solutions
Level 1
A point P is 0.50 meter from a point charge of 5.0 × 10 –8 coulomb.
1. The intensity of the electric field at point P is most nearly
(A) 2.5 × 10–8 N/C (B) 2.5 × 101 N/C (C) 9.0 × 102 N/C (D) 1.8 × 103 N/C
(E) 7.5 × 108 N/C
E = kQ/r2
2. The electric potential at point P is most nearly
(A) 2.5 × l0–8 V (B) 2.5 × 101 V (C) 9.0 × 102 V (D) 1.8 × 103 V (E) 7.5 × 103 V
V = kQ/r
Level 2
Two charges are located on the line shown in the figure below, in which the charge at point I is +3q and the
charge at point III is +2q. Point II is halfway between points I and III.
3. Other than at infinity, the electric field strength is zero at a point on the line in which of the following ranges?
(A) To the left of I (B) Between I and II (C) Between II and III (D) To the right of III
(E) None; the field is zero only at infinity.
Where E is zero must be closer to the smaller charge to make up for the weaker field. The
vectors point in opposite directions when between the two like charges. These two criteria
eliminate 4 of the choices
4. The electric potential is negative at some points on the line in which of the following ranges?
(A) To the left of I (B) Between I and II (C) Between II and III (D) To the right of III
(E) None; this potential is never negative.
Since both charges are positive and V is a scalar equal to ΣkQ/r, the potential will never be zero
in the vicinity of these two charges.
5. Points R and S are each the same distance d from two unequal charges, +Q and +2Q, as shown above. The work
required to move a charge –Q from point R to point S is
(A) dependent on the path taken from R to S
(B) directly proportional to the distance between R and S
(C) positive
(D) zero
(E) negative
By symmetry VR = VS so ΔVRS = 0 and W = qΔV
Free Response
2005Bb3 The figure above shows two point charges, each of charge –2Q, fixed on the y–axis at y = +a and
at y = –a. A third point charge of charge –Q is placed on the x–axis at x = 2a. Express all algebraic
answers in terms of Q, a, and fundamental constants.
a. Derive an expression for the magnitude of the net force on the charge –Q due to the other two charges,
and state its direction.
b. Derive an expression for the magnitude of the net electric field at the origin due to all three charges,
and state its direction.
c. Derive an expression for the electrical potential at the origin due to all three charges.
d. On the axes below, sketch a graph of the force F on the –Q charge caused by the other two charges as
it is moved along the x–axis from a large positive position to a large negative position. Let the force be
positive when it acts to the right and negative when it acts to the left.
Solution
Free Response 2
2001B3. Four charged particles are held fixed at the corners of a square of side s. All the charges have the
same magnitude Q, but two are positive and two are negative. In Arrangement 1, shown above, charges
of the same sign are at opposite corners. Express your answers to parts a. and b. in terms of the given
quantities and fundamental constants.
a. For Arrangement 1, determine the following.
i. The electrostatic potential at the center of the square
ii. The magnitude of the electric field at the center of the square
The bottom two charged particles are now switched to form Arrangement 2, shown above, in which the
positively charged particles are on the left and the negatively charged particles are on the right.
b. For Arrangement 2, determine the following.
i. The electrostatic potential at the center of the square
ii. The magnitude of the electric field at the center of the square
c. In which of the two arrangements would more work be required to remove the particle at the upper
right corner from its present position to a distance a long way away from the arrangement?
_________ Arrangement 1 ___________ Arrangement 2
Justify your answer
Solution
Jedi
This guy is difficult. Let me know if you find an easy way to solve it. I looked for an easier way with no luck.
6. Two charges, –2Q and +Q, are located on the x–axis, as shown above. Point P, at a distance of 3D from the
origin O, is one of two points on the positive x–axis at which the electric potential is zero. How far from the
origin O is the other point?
(A) 2/3 D (B) D (C) 3/2 D (D) 5/3 D (E) 2D
The points where V = 0 must lie closer to the smaller charge. Unlike electric field vectors which
also require the individual vectors point in opposite directions, there are a locus of points (in this
case in a ring surrounding the +Q charge) where V = 0 as the two charges are opposite in sign
and V is a scalar. So the other point on the x axis is between the two charges, but closer to the
+Q charge. This must be a value between 1.5 D and 2 D
Answer: A
By symmetry, all the vectors cancel
Answer: D
W = qΔV = +Q(Vcenter – V∞) = +QVcenter where Vcenter = ΣV = ΣkQ/r = 6kQ/R