Electric Potential and Electric Potential Energy: What's the difference?

1. SIR JOHNEL V.
ESPONILLA
GENERAL PHYSICS II

2.  Relate the electric potential with work, potential energy, and electric
field.
 Determine the electric potential function at any point due to highly
symmetric continuous charge distributions.
 Infer the direction and strength of electric field vector, nature of the
electric field sources, and electrostatic potential surfaces given the
equipotential lines.
 Define the work done by the electric force.
 Calculate the electric field in the region given a mathematical
function describing its potential in a region of space.
 Solve problems involving electric potential energy and electric
potentials.

5. DIRECTIONS: Check what you have learned about electric fields and charges in
the previous lesson. Write CHARGE if the statement is correct; otherwise write
FIELD in your paper.
1. The SI unit of electric charge is the coulomb, symbol
C.
2. Electrostatic effects occur when electrical charges
are separated.
3. All charge separation involves the expenditure of energy.
4. The magnitude of the electrostatic force between
charges increases as their separation decreases.
CHARG
E
CHARG
E
CHARG
E
CHARG
E

6. POTENTIAL ENERGY -----
GRAVITATIONAL

7. If we want to move the test
charge to plate B, a force
must be applied to push
against the force of the

8. If the force applied is
conservative, the work
done can be expressed as
POTENTIAL ENERGY U.

9. If Wab is positive, Ua is greater than Ub, ∆U is negative,
potential energy decreases.
Whether the test charge in the e-field is positive or negative,
the potential energy increases if the test charge moves
OPPOSITE to the direction of the electric force.
The potential energy decreases if the test charge
moves in the SAME direction of the electric force.

10. where r is the distance of separation, qqo are the values of two
charges, electric potential energy is expressed as Nm or Joules
(J)
For electric potential energy with several test
charges, this is expressed as

11. A point charge q1 = +2.80𝜇C is at origin. How far
should the second point charge of +5.20 𝜇C be placed
to have electric potential energy of 0.600 J?

12. A point charge q1 = +2.80𝜇C is at origin. How far
should the second point charge of +5.20 𝜇C be placed
to have electric potential energy of 0.600 J?
given
unknown sketch
Q1 = +2.80 x 10-6 C ;
Q2 = +5.20 x 10-6 C ;
U = 0.600 J
r

13. A point charge q1 = +2.80𝜇C is at origin. How far
should the second point charge of +5.20 𝜇C be placed
to have electric potential energy of 0.600 J?
strategy
solution
Rearrange the formula and solve for r
since it is unknown. Then, plug-in for
the values.
𝑟 =
1
4𝜋𝜖0
𝑞𝑞0
𝑈
r =
1
4𝜋𝜖0
(2.80 𝑥10−6𝐶)(5.20𝑥10−6𝐶)
0.600 𝐽
r = 9 x 109 𝑁𝑚2
𝐶2
1.456𝑥10−11𝐶2
0.600 𝑁𝑚
r = 0.2184 m

14. A charge of 4.50 x 10-8C is placed in a uniform electric field
that is directed vertically upward with a magnitude of
5.00 x 104 N/C. What work is done by the electrical force
when the charge moves 0.450 m to the right? 0.800 m
downward? 2.60 m at an angle of 45 degrees from the
horizontal?
given
unknown sketch
Q1 = +4.50 x 10-8 C ;
E = 5.00 x 104 N/C ;
r = 0.450 m, r = 0.800
m
r = 2.60 m, 450
W

15. A charge of 4.50 x 10-8C is placed in a uniform electric field
that is directed vertically upward with a magnitude of
5.00 x 104 N/C. What work is done by the electrical force
when the charge moves 0.450 m to the right? 0.800 m
downward? 2.60 m at an angle of 45 degrees from the
horizontal?
strategy
solution
Use W = Fd cos𝜃, where F = qE.
W = Fd = qEd = (4.50 x 10-8 C)(5.00 x 104
N/C)(0.45 m)(cos90) = 0J
W = Fd = qEd = (4.50 x 10-8 C)(5.00 x 104
N/C)(0.800 m)(cos0) = -1.8 x 10-3 J
W = Fd = qEd = (4.50 x 10-8 C)(5.00 x 104
N/C)(2.60 m)(cos45) = 4.137 x 10-3 J

16. where U is the electric potential energy and q0 is the
charge, electric potential is expressed as volt (V)
Divide both sides of the equation relating work done
by the electric force from points a to b by q0 to
represent it as work per unit charge.

17. The difference Va – Vb is known as the potential difference.
To find the electric
potential V of a
single point charge,
express as
1. The potential at any point is
negative if q is negative.
2. The potential at any point is
positive if q is positive.
3. The potential is zero at r =
infinity.
To find the electric
potential V of a
severe point
charges, express as
It is the scalar sum of the fields
produced by each charge.

18. A point charge has a charge of 8.00 x 10-11 C. At what
distance from the point charge is the electrical
potential 12.0 V? 24.0 V?
given
unknown sketch
Q1 = +8.00 x 10-11 C ;
V1 = 12.0 V ;
V2 = 24.0 V
r

19. A point charge has a charge of 8.00 x 10-11 C. At what
distance from the point charge is the electrical
potential 12.0 V? 24.0 V?
strategy
solution
Re-arrange the formula and solve for r
since it is unknown, then plug-in the
values.
𝑟 =
1
4𝜋𝜖0
𝑞
𝑉
r =
1
4𝜋𝜖0
(8.00 𝑥10−11𝐶)
24 𝑉
r = 9 x 109 𝑁𝑚2
𝐶2
3.33 𝑥 10−12𝐶
𝐽/𝐶
r = 0.03 m at 24 V

20. A point charge has a charge of 8.00 x 10-11 C. At what
distance from the point charge is the electrical
potential 12.0 V? 24.0 V?
solution 𝑟 =
1
4𝜋𝜖0
𝑞
𝑉
r =
1
4𝜋𝜖0
(8.00 𝑥10−11𝐶)
12 𝑉
r = 9 x 109 𝑁𝑚2
𝐶2
6.67 𝑥 10−12𝐶
𝐽/𝐶
r = 0.06 m at 12 V

21. DO
THIS!!!
In a certain region an electric potential V is present. An
unknown charge Q is moved around this region between
points at potential difference ∆V. A physicist measures the
change in the potential energy ∆U as the charge is moved.
The table shows the data of the measurements. The unit
for ∆V are MV (mega-Volts), the units for ∆U are kJ (kilo-
joules).
∆𝐕 52.1 24.9 11.7 4.97 3.55 3.32 1.52 1.87
∆𝐔 357.8 152.6 119.3 43.8 48.9 43.8 32.9 42.3