Physics electric something

1. General Physics 2
Activity Sheet
Quarter 3 – MELC 15 &17
Week 3 Electric
Potential Function
Background Information for Learners

2.  Earth’s gravitational force is somewhat analogous to
electric potential. This is so because, a moving positive test
charge against the direction of an electric field is moving a
mass upward within Earth’s gravitational force. Their
movements would be like going against nature and work
done by an external force. This means that work increases
its potential energy of the object. Likewise, the movement
of a positive test charge in the direction of an electric field
would be like a mass falling downward within Earth’s
gravitational field. Thus, this motion would result in the
loss of potential energy.
 In the same manner, gravitational potential is location-
dependent, and is independent on the mass of the object
experiencing the field. It describes the effects of
gravitational field in objects that are place on the location
within it.

3.  To further understand the relationship exhibited by
the electric potential between potential energy,
work and electric field, consider the figure below
The electric field created by a positively charged Van
de Graaff generator. The illustration in the figure
shows that the direction of the electric field is in the
direction that a positive charge would be pushed; for
this case, the direction is outward and away from the
Van de Graaff. Work is move to a positive test charge
towards the sphere against the electric field and the
amount of force involved in doing the work is
dependent upon the amount of charge being moved.
(according to Coulomb’s Law of Electricity). Thus,

4. work would change the potential energy by an
amount that is equal to the amount of work done.
 Furthermore, the electric potential energy is
dependent upon the amount of charge on the object
experiencing the field and upon the location within the
field. Electric potential is the potential energy per
charge. This concept was used to express the effect of
an electric field to a source based on the location
within the electric field. There should be a positive
test charge at a higher electric potential when it held
closed to a positive source and at a lower electric
potential when it’s
held farther. In this manner, electric potential
becomes

5.  dependent of the location within an electric field.
Therefore, the relationship between potential and
field is a differential, that, as the test charge was
moved in x direction, the rate of its change in potential
was the value of electric field.
Activity 1: Let’s volt in!
A. Potential energy per unit charge
The potential at any point of an electrostatic field is
also known as potential energy per unit charge at the
point Mathematically, V = U/q’ or U = q’V
Where:
V=potential
q’= charge
U= potential energy

6. • Potential energy and charge are both scalars
• SI unit is 1 joule per coulomb (1J/c)
• 1J/C is equal to 1 volt
B. On a work per unit charge basis
In equation: Wa-b= Ua – Ub = Va - Vb
q’ q’ q’
where:
Va = Ua/q’ is the potential per unit charge at point a
Va & Vb = are called the potential at point a and
point b

7. C. Potential energy U of the test charge q’ at any
distance r from charge q is given by: In equation: U = 1
U = 1 qq’
4πЄo r
D. The potential V at any point due to an
arbitrary collection of point
charge is given by:
In equation V= U/q’ = 1 Σ qi/ri
4πЄ0
Where:
1 , is the potential energy when q is at point a, at a distance r from q
4πЄo.

8. V= potential at any point in electrostatic field
r = Distance from respective charges from point a to point b.
q=point charge distribution
Note that potential is like electric field which is independent
of the test charge q’ used to define it.
• When a charged particle moves in a region of space
where there is an electric field, the field exerts a force,
and does work on it. One simple example is a pair of
charged parallel metals set up a uniform electric field of
magnitude E in the region between them, and the
resulting force on a test charge q’ has a magnitude
F= q’E.

9. •When the charge moves from one point to another point,
the work done by this force on the test charge is Wa-b= q’Ed
(This work was represented by a potential energy function).
• If you take the potential energy to be zero at point b,
then at point a it has a value q’Ed, thus at any point a
distance y above the bottom plate is given by U(y) = q’Ey
•And when the test charge moves from height y1 y2,
the work done by the field is given by
W 1-2= U (y1) – U (y2) = q’ Ey1 – q’Ey2
• When y1 is greater than y2, U decreases and the field does
positive work
• When y1 is less than y2, U increases and the field does
negative work

10. Sample Problem:
1. A particle having a charge q= 3x10-⁹
c moves from
point a to point b along a straight line, with a total
distance of d=0.5m. The electric field is uniform along
this line, in the direction from a to b, with magnitude,
E = 200 N/C. Determine the force on q, the work done
on it by the field, and the potential difference Va – Vb.
Given:
q’ = 3x10-
⁹C
E = 200 N.C-¹
Solution:
The force is in the same direction as the electric
field, and its magnitude, thus in
a.F = q.E = (3X10-
⁹C) (200 N.C-1
) = 6 x 10-7
N

11. The work done by this force is
b) W = F. d = (6x10-7
N) (0.5m) = 3 x10-7
J
c) Va – Vb = W/q = 3 x 10-7
J / 3 x10-⁹
C = 100
J.C-¹
= 100 V
Va – Vb = E.d = (200 N.C-¹)
(0.5m) = 100 J. C-¹
=
100 V
(Note. You can either use any of the two equation in
solving for potential difference).
2. Point charges of + 12 x10-⁹
C and -12x10-⁹
C are placed 10 cm
apart. Compute the potentials at points a and b, given that the
distance at point a from the positive charge is 6cm while on
the negative chrage.is 4cm. And at point b, the distance from
the positive charge is 4cm and 14 cm on the negative charge.

12. Given:
q+ = 12 x 10-⁹
C
q- = 12 x 10-⁹
C
da = 6cm(q+),4cm (q-)
db = 4cm (q+), 14cm (q-)
Solution:
Using the equation to evaluate the algebraic sum (1/4πЄ0 Σ
qi/ri )
The potential: at point a due to a positive charge is
9.0 x10⁹N.m². C-
²) 12 x 10-
⁹C = 1800 N.m. C-
¹= 1800 V
0.06 m

13. at point a due to a negative charge is
(9.0 x10⁹N.m² C-
²) -12 x 10-
⁹C = - 2700 N.m. C-
¹ = -2700 V
0.04 m
Thus, Va = 1800V -2700 V = -900 V = -900 J.C-
¹
At point b due to a positive charge is
9.0 x10⁹N.m². C-
²) 12 x 10-
⁹C = 2700N.m. C-
¹ = 2700 V
0.04 m
At point b due to a negative charge is
(9.0 x10⁹N.m². C-
²) -12 x 10-
⁹C = - 770N.m. C-
¹ = -770 V
0.14 m
Thus, Vb = 2700 V -770 V = 1930 V the potential

14. 3. Compute the potential energy of a point charge of +
4 x 10 -9 C if placed at points a and b in problem 2.
Given:
q+ = 4 x 10-
⁹C
Va = -900 V
Vb = 1930 V
Solution:
Using the formula U= qV
at point a
U = (4 x10-
⁹C) (-900 J. C-
¹) = -36 x10 -
⁷J at
point b

15. U = (4 x10-
⁹C) (1930 J. C-
¹) = 77 x10 -
⁷J
Solve the following problems:
1. The potential at a certain distance from a point
charge is 600 V, and the electric field is 200 N.C-
¹.
What is the distance to the point charge?
2. A potential difference of 2000 V is established
between parallel plates in air. If the air becomes
electrically conducting when the electric intensity
exceeds 3x10⁶N.C-¹
, what is the minimum
separation of the plates?
3. A particle having a mass m= 5g and charge q’=
2x10-⁹C starts from rest at point a and moves in a
straight line to point b, under the influence of the

16. electric fields of the two charges. What is its speed
v at point b?
Activity 2:
1.If the electric potential at a single point is known can the
electric field at point be determined? 2.Why is potential
gradient a scalar quantity?
3.How would you respond to this statement: “Since electrical
potential is
always proportional to potential energy, why bother with the concept
of potential at all?
4. Why it is easy to produce a potential of several
thousand volts on your body by scuffing your shoes across
a nylon carpet than with a contact in a power line?

17. Reflection
What insights about the lesson that will help you in using the
appliances at your home? Cite examples that will concretely
shows your learnings.