Electric Fields:

Electric Charge

The phenomena of two objects sticking together can be explained by the notion that objects when rubbed can gain a net electric charge. There are two types of charge, labeled positive ( + ) and negative ( - ), with the following basic property:

Like charges of the same sign repel each other.
Unlike charges of the opposite sign attract each other.

Detailed experiments have established the following fundamental characteristics of electric charge:

Charge is never created nor destroyed - it is conserved.
Charge always comes in an integral multiple of a basic unit - it is quantized.

This basic unit of charge is conventionally denoted by e :

e = 1.602 x 10^{- 19} Coulombs (C) .

(1)

In an atom, the charge on an electron is - e and that on a proton is + e .

Electric Forces and Fields

The amount of attraction or repulsion between charged objects can be put in quantitative terms by the introduction of the electric force. The simplest case to consider is the force between two point charges (charges with a negligible size). Experiments by Coulomb and others uncovered the following formula for the force $\vec{F}$ between two such charges Q₁ and Q₂ separated by a distance r , as in Fig. 1.1:

Figure 1.1: Force between two point charges Q₁ and Q₂

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The magnitude of the force between the charges is

F = k

(2)

where k is a constant called Coulomb's constant:

k = 9.0 x 10⁹

(3)

The direction of the force is along a line joining the two charges. It is repulsive if the charges have the same sign and is attractive if the charges have opposite sign.

If there are more than two charges present, then the force on any one charge must be found by adding vectorially the forces found by Coulomb's law (1.2) between each pair of charges.

It is convenient for many applications to introduce the concept of the electric field, conventionally denoted by $\vec{E}$ . Suppose we have a ``background'' distribution of charge Q₁,Q₂,...,Q_n in some region of space, and measure the force $\vec{F}$ on a charge q placed nearby. The electric field $\vec{E}$ associated with this charge distribution is defined through the relation

$\displaystyle\vec{F}$ = q $\displaystyle\vec{E}$ .

(4)

The units of $\vec{E}$ are thus seen to be N/C. In a sense, the charge q is a test charge which probes the strength at different points of the potential electric force due to the charges Q₁,Q₂,...,Q_n . Note that, for a given electric field $\vec{E}$ , the force on a positive charge is opposite in direction to the force on a negative charge.

For a single point charge Q the electric field a distance r away is found from Eqs.(1.2,1.4) to have the magnitude

E = k Q/r²,

(5)

with a direction equal to the direction of the force on a positive test charge placed at the point of interest. As with the electric force, the electric field due to multiple point charges must be found by adding vectorially the electric field found by Eq.(1.5) for each individual charge.

Constant Electric Field

We begin by considering a constant electric field similar to one found between the two plates of a parallel plate capacitor. This is analogous to the uniform gravitation field found close to the surface of a flat Earth.

The Electric Field lines, point in the direction of the force that a positive would experience (upwards). The force acting on a charge Q is F = QE, and the size and direction is constant within the space between the two plates. The graph for a uniform field would be:

When a charge moves from one position to another in the direction of the Force the work done can be calculated from the area between the graph and the horizontal axis.

The energy gained, W = F(y₂ – y₁) = QE (y₂ – y₁).

If we were to define the electric potential energy within this field we would need to specify a reference position, which would normally be the positive plate. For a charge, Q placed at a position y above the positive plate the energy it would gain if it moved to that plate would be W = - QEy. The negative sign comes about because when the charge is positive it is being repelled from the positive plate and as such loses energy in moving to the plate. (The formula will work for a negative charge as well, when we substitute the value of the charge and its sign we will obtain a positive value for W, which is correct because the charge would be attracted to the positive plate.) The electric potential energy in a uniform electric field is given with respect to the positive plate and equals:

E_p = -QEy

Be careful of the use of the two Es, the first one is the electrical potential energy and the second one on the RHS of the equation is the Electric Field Strength.

Note that we would have obtained the electric potential energy by calculating the integral:

which would have turned out to be negative.

The electric potential (V) at a point is defined as the electric potential energy that a +1 C of charge would have at that point. This would give us that V = -Ey for the case of a uniform field.

Electric Field due to a single charge

The equation of how the Electric Field varies with position is given by

F = k

Where Q is the charge that provides the source of the Electric Field found at a position r from it. The field is radial and points along the line between Q and the point directed away from Q when Q > 0 and towards Q when Q < 0.

The graph of the variation of E with r is shown below.

To calculate the Work Done on a charge when it changes position in the field we need to calculate the area under the graph, taking care that we note the direction of the change.

For Calculus we would calculate,

W =

When it comes to defining the electrical potential energy in such a field the point at infinity is taken as the reference position. The electric potential is then the area represented by the area between the point and and infinity

This is the integral:

which gives, V =

Note that this integral gives us a positive value for the electric potential at a point when Q > 0 and a negative value when Q < 0.

Hence the electrical potential energy for a charge q is given by k. Whether this value is positive or negative depends on the signs of the two charges.

Problems

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Three point charges, q₁ = - 4 nC, q₂ = 5 nC, and q₃ = 3 nC, are placed as in Fig. 1.4.

Figure 1.4: Three point charges

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If r₁ = 0.5 m and r₃ = 0.8 m, find the force on q₂ due to the other two charges.

Solution:
We first find the force on q₂ due to q₁ :

F₁ = k $\displaystyle{\frac{q_1q_2}{r_1^2}}$ = 9.0 x 10⁹ x $\displaystyle{\frac{4\times 10^{-9}\cdot 5\times 10^{-9} }{(0.5){}^2}}$ = 7.2 x 10^-
7 N ,

which is directed to the left. The force on q₂ due to q₃ is found as:

F₃ = k $\displaystyle{\frac{q_3q_2}{r_3^2}}$ = 9.0 x 10⁹ x $\displaystyle{\frac{3\times 10^{-9} \cdot 5\times 10^{-9} }{(0.8){}^2}}$ = 2.11 x 10^-
7 N ,

which is also directed to the left. Thus, the total force on q₂ is given by

7.2 x 10^-
7 + 2.11 x 10^{- 7} = 9.31 x 10^-
7 N .

Thus, the total force on q₂ is 9.31 x 10^{- 7} N directed to the left.

PROBLEM 1.2

A m = 2 g ball is suspended by a l = 20 cm long string as in Fig. 1.6 in a constant electric field of E = 1000 N/C. If the string makes an angle of $\theta$ = 15^owith respect to the vertical, what is the net charge on the ball?

Figure 1.6: Charged ball in an electric field

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Solution:
We first examine in Fig. 1.7 the forces present on the ball.

Figure 1.7: Force diagram

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In equilibrium the net force on the ball is zero; we thus have, in the y direction,

Tcos $\displaystyle\theta$ - mg =

0 $\displaystyle\Rightarrow$

T =

$\displaystyle{\frac{mg}{\cos\theta}}$

and in the x direction,

qE - Tsin $\displaystyle\theta$ =

0 $\displaystyle\Rightarrow$ q =

$\displaystyle{\frac{T\sin\theta}{E}}$

$\displaystyle{\frac{mg\tan\theta}{E}}$ .

With the numbers given, we find

q =

$\displaystyle{\frac{0.002\cdot 9.8 \cdot\tan\,15^\circ}{1000}}$

= 5.2 x 10^{- 6} C .

Thus, the charge on the ball is 5.2 μC.

PROBLEM 1.3

The three charges in Fig. 1.8, with q₁ = 8 nC, q₂ = 2 nC, and q₃ = - 4 nC, are separated by distances r₂ = 3 cm and r₃ = 4 cm. How much work is required to move q₁ to infinity?

Figure 1.8: Three point charges

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Solution:
We will first calculate the electric potential V due to a point charge q a distance r away as

V = k q/r

which assumes the potential vanishes at infinity. The potential due to q₂ at the point occupied by q₁ is thus

V₂ = k q₂/r₂= 9.0 x 10⁹ x (2 x 10^-9 )/0.03= 600 V ,

while that due to q₃ is

V₃ = k q₃/r₃ = 9.0 x 10⁹ x (-4 x 10^-9)/0.05= - 720 V ,

The net potential is thus

V = 600 + (- 720) = - 120 V ,

and so the work required to move q₁ to infinity is

W = - q₁ $\displaystyle\Delta$ V = - q₁ x (V_f - V_i) = - 8 x 10^-
9 x (0 – (-120) ) = - 9.6 x 10^-
7 J .

Thus, the work required is 9.6 x 10^-
7 J, with the minus sign indicating that a net work must be supplied.