電位 / Electric Potential

Going to do three questions today related to electric potential (More commonly known as Voltage) and electric forces.


1) Suppose we have two negatively charged plates placed next to each other an arbitrary distance "d" away from each other. What is the electric field at points 1, 2, and 3?

Recall the formula for electric field for a charged plate:
|E| = |σ/2ε| where σ = Q/A and ε is the permittivity value in a vacuum: 8.85E-12


At point 1, a test charge (which is a positive point charge with negligible charge) will be attracted to both plates (Because positive charges are attracted to negative ones). Therefore, the total electric field are additive.


σ/2ε + σ/2ε = σ/ε

From this result, we can see that the distance the plates are from each other does not matter.


At point 2, a test charge would once again be attracted to both plates, but since the attractive forces act in opposite directions, they cancel each other out.


σ/2ε - σ/2ε = 0

Finally, at point 3, we get the same result as point 1, in the case that the test charge is attracted to both plates. 

σ/2ε + σ/2ε = σ/ε


2) An object of mass 1.0 g is attached to a 50 cm string and hangs from a positively charged plate. The plate exerts an electric field of 10,000 N/C, and the object on the string is at a new equilibrium position. What is the charge, q, of the object?

We first draw a free body diagram of the object at its new equilibrium position:

We can see that three forces act on the object: Electric force, gravitational force, and tension on the string.

Recall the two Laws of Equilibrium: All forces acting on an object must equal 0, and all torques acting on the same object must equal 0. Since there is no circular motion involved, we only need to focus on the first law.


ΣF_e = 0
ΣF_e = ΣF_ex + ΣF_ey = 0

What is F_e? We can relate F_e and E by the equation:

E = F/q
F = E*q 

What is F_g? This is a simple application of Newton's Second Law (F = ma) 

F_g = m * g


What is T? T is composed of the horizontal and vertical vector components.


T = Tx + Ty
Tx = T*sin(θ)
Ty = T*cos(θ)

We can calculate θ with simple trigonometry:

θ = cos^-1(49/50) = 11.48

We can now go back to our equilibrium problem:

ΣF_ex = 0
|Tx| = |Fe|
T*sin(θ) = E*q


ΣF_ey = 0
|Ty| = |m*g|

T*cos(θ) = m*g


Since we have two unknowns and two equations, we can substitute one to solve the other.


T = m*g/cos(θ)



Tsin(θ) = E*q
(m*g/cos(θ))(sin(θ)) = E*q
m*g*tan(θ) = E*q
q = (m*g*tan(θ))/ E

Finally solve by plugging in appropriate values:


q = ((0.001 kg)(9.8 m/s^2)(tan(11.48))) / (10,000 N/C)
q = 1.99E-7 C


3) Suppose we have two point charges, q1 = 7.5 C, q2 = -2.5 C, that are 5 cm and 8 cm from the origin respectively. Where the x-axis, to the right of q2, is the electric potential zero?

Recall the formula for electric potential:

V = kq/r

We want to calculate the potential difference such that it is equal to zero:


V = kq1/r1 + kq2/r2 = 0
V = k(7.5)/(3+x) + k(-2.5)/x = 0

7.5/(3+x) = 2.5/x
7.5x = 7.5 + 2.5x
5x = 7.5
x = 7.5/5 = 1.5 cm


Therefore, the electric potential is zero at a distance of 5 + 3 + 1.5 = 9.5 cm from the origin.