Coulomb's Force

COULOMB’S FORCE LAW Two point charges Multiple point charges Attractive Repulsive

- - + + +

The force exerted by one point charge on another acts along the line joining the charges. It varies inversely as the square of the distance separating the charges and is proportional to the product of the charges. The force is repulsive if the charges have the same sign and attractive if the charges have opposite signs. q1 Two point charges q1 and q2

q2 [F]-force; Newtons {N} [q]-charge; Coulomb {C} origin [r]-distance; meters {m} [ε]-permittivity; Farad/meter {F/m} Property of the medium

COULOMB FORCE UNIT VECTOR Permittivity is a property of the medium. Also known as the dielectric constant.

Permittivity of free space

Coulomb’s constant Permittivity of a medium Relative permittivity

For air

FORCE IN MEDIUM SMALLER THAN FORCE IN VACUUM Insert oil drop

Viewing microscope Eye Metal plates

Millikan oil drop experiment

Charging by contact

Example (Question):

A negative point charge of 1µC is situated in air at the origin of a rectangular coordinate system. A second negative point charge of 100µC is situated on the positive x axis at the distance of 500 mm from the origin. What is the force on the second charge? Example (Solution):

q1 = -1 µC origin X q2= -100 µC Example (Solution): Y

q1 = -1 µC origin X q2= -100 µC

END Multiple point charges

It has been confirmed experimentally that when several charges are present, each exerts a force given by

on every other charge. The interaction between any two charges is independent of the presence of all other charges. 5 point charges: Net force on q5

q3 Example (Question) The charging of individual raindrops is ultimately responsible for the electrical activity in thunderstorms. Suppose two drops with equal charge q are located on the x axis at ± a. Find the electric force on a third drop with charge Q located at an arbitrary point on the y axis. Example (Solution) The charging of individual raindrops is ultimately responsible for the electrical activity in thunderstorms. Suppose two drops with equal charge q are located on the x axis at ± a. Find the electric force on a third drop with charge Q located at an arbitrary point on the y axis. Y Q

y q q X a a Example (Solution) Y Q

y q q X a a Charge Q is the same distance r from the two other charges, so the force from each has the same magnitude:

The direction of the two forces are different. Y Example (Solution)

The x components cancel, Q while the y components add.

y q q X a a Example (Solution) Y From figure

Then Q

y q q X a a Example (Solution) Y

Consider the case where the charge is also located on the x axis, y = 0.

Q q Q q a a X y q q X a a Example (Solution) Y

Consider the case where Y the charge is a very large distance on the y axis such that y >> a. Q Q

y q q X a a (2q) END Electric field

Single point charge Multiple point charges Charge distribution We define an electric field similar to that of a gravitational field.

A charge produces an electric field such that when another “test” charge is placed in the field it will experience an electrical force.

- Electric field

Test charge

Electric force The electric field at any point is the force per unit charge experienced by a charge at that point.

Since the electric force is a vector then the electric field is also a vector. Electric field and electric force are vectors which point in the same direction when the test - charge q is positive.

The electric field lines for a positive source charge point away from the source charge.

- The electric field lines for a negative source charge point towards the source charge. Consider two point charges (+Q and +q) again and the force that exists between them.

Electric force on q

Electric field at q Q P Observation point

Q [k]-Coulomb constant; meter/Farad {m/F} [q]-charge; Coulomb {C} [r]-distance; meters {m} [E]-Electric field; Newton/Coulomb {N/C} [E]-Electric field; Volt/meter {V/m}

ELECTRIC FIELD (1) Electric field is a vector quantity. Thus at all points where the electric field exists it has magnitude and direction. (2) The charge q must be small and positive such that it does not disturb the source charge Q.

(3) For a positive source charge Q the electric field vector and the electric force on the test charge q are in the same direction. (4) For a positive source charge Q, the electric field lines are directed away from the charge. (5) For a point charge Q located at the origin the electric field vector is: Y

Q + X

Example (Question) What is the electric field 30 cm from a charge q = 4.0 nC? Example (Solution) What is the electric field 30 cm from a charge q = 4.0 nC?

Z P

q Y

X Example (Alternate Solution) Obtain the force on a test charge

q Y

X Then obtain the electric field END Electric field produced by multiple point charges (n = 2)

q1 Z P q2

Y q1 and q2 are the source point charges. X P is the field point

The charges (q1 and q2) produce the electric field observed at the point P Consider each charge in turn, independently of all other charges present.

q1 Z P q2 Charge q1 produces an electric field at point P. Y Charge q2 produces an electric field at point P

X The total electric field at P is the vector sum of the electric field produced by each individual charge. Consider charge q1 only

q1 Z P q2

Electric field produced by charge q1 at P Y

Distance separating q1 and P

Unit vector along line joining q1 and P Consider charge q2 only

q1 Z P q2

Electric field produced by charge q2 at P Y

Distance separating q2 and P

Unit vector along line joining q2 and P The total electric field at P is the vector sum of the electric field produced by each individual charge.

q1 Z P q2

X Electric field produced by multiple point charges (n > 1)

q1 q3 qi Z P q2 q q 4 The total electric field at P is the n vector sum of the electric field Y produced by each individual charge.

X q5 Given a group of charges we find the net electric field at any point in space by using the principle of superposition. This is a general principle that says a net effect is the sum of the individual effects. Here, the principle means that we first compute the electric field at the point in space due to each of the charges, in turn. We then find the net electric field by adding these electric fields vectorially, as usual. Example (Question) The figure shows two point charges each of +10 nC separated in air by 8.0 m. Compute the electric field at the points A, B, and C

+ + Example (Solution) The figure shows two point charges each of +10 nC separated in air by 8.0 m. Compute the electric field at the points A, B, and C Point A: Make a sketch of the layout and then draw in vectors for

the fields E1 produced by q1 and E2 produced by q2. To do that imagine a positive test charge at A. The force on it due to the charge

q1 acts along the center-to-center line, is repulsive, and so points to the right. That means the E1 at A is to the right along the axis. Similarly, the force due to q2 on our imaginary test charge is to the left as is E2. Next calculate E1 and E2 and add them vectorially. We are spared this effort since E1 = E2, the two cancel and the field at A is zero

+ + Example (Solution) The figure shows two point charges each of +10 nC separated in air by 8.0 m. Compute the electric field at the points A, B, and C

Point B: At point B the fields act as drawn in the figure, and we must find their components. First we will calculate E1 and E2.

+ + X Example (Solution) Point B: Since the charges and distances happen to be the same, the magnitudes of the two contributing fields are equal:

Now for the vector components

+ + X Example (Solution) Point B: Since the charges and distances happen to be the same, the magnitudes of the two contributing fields are equal:

 N E1 = 2.81 (cos(45)xˆ + sin(45)yˆ ) ( C)  N E 2 = 2.81 (−cos(45)xˆ + sin(45)yˆ ) ( C) €

€

+ + X Example (Solution) Point B: The horizontal field components are equal and act in opposite direction. They will cancel. Only the vertical field components contribute, and in the same direction.

N E = E sin 45o + E sin 45o = 2(2.81 )(0.707) B 1 ( ) 2 ( ) C

€ Y

The direction is straight up in the positive y-direction + + X Example (Solution) Point C: The point C is similarly located with respect to the charges as point B is.The field magnitude at C is the same as at B except the direction is straight down in the negative y-direction.at

The direction is + + straight down in the negative y-direction Example (Solution) The figure shows two point charges each of +10 nC separated in air by 8.0 m. Compute the electric field at the points A, B, and C Additional figures related to this example question

+ +

END In is often useful to imagine Charged line that there is a continuous distribution of charge.

The principle of superposition Charged surface applies provided the charge distribution is divided into small elements of charge Δq and the t o t a l e l e c t r i c f i e l d a t a n observation point is obtained by Charged volume summing all electric field contributions from each element Δq. In is often useful to imagine Charged line that there is a continuous distribution of charge.

Charged surface Recall in calculus

Charged volume The electric field at the point P is the sum of the vectors arising from the individual charge elements dq in the entire distribution, each calculated using the appropriate distance r and unit vector . P Electric field produced at P by one segment dq

Electric field produced by all segments along line of length L

Linear charge density on the line Units; {C/m}

Charge on length segment P

may be a function of the coordinates usually a constant

Integration over length of line charge unit vector function of (x,y,z),…. P Electric field produced at P by one segment dq

Electric field produced by all Surface charge density segments of surface S

Units; {C/m2}

Charge on surface element dA P

ρs may be a function of the coordinates usually a constant

Integration over surface of charge unit vector function of (x,y,z),…. P Electric field produced at P by one segment dq

Electric field produced by all segments in volume V V Volume charge density

Units; {C/m3}

Charge in volume element dV P

ρV may be a function of the coordinates usually a constant

Integration over volume of charge unit vector function of (x,y,z),…. Example (Question) W i r e s , a n t e n n a s , a n d s i m i l a r elongated structures can often be considered as thin rods carrying electric charge. Suppose a rod of length carries a positive charge Q distributed uniformly over its length. Find the electric field at the point P a distance a from the end of the rod. Example (Solution) W i r e s , a n t e n n a s , a n d s i m i l a r elongated structures can often be considered as thin rods carrying electric charge. Suppose a rod of length carries a positive charge Q distributed uniformly over its length. Find the electric field at the point P a distance a from the end of the rod.

Let the y axis lie along the rod, with origin at P. Consider a small length dy of the rod, containing charge dq, and located a distance y from P. A unit vector from dq to P is:

The field at P due to dq is: Example (Solution) W i r e s , a n t e n n a s , a n d s i m i l a r elongated structures can often be considered as thin rods carrying electric charge. Suppose a rod of length carries a positive charge Q distributed uniformly over its length. Find the electric field at the point P a distance a from the end of the rod.

The net field at P is the sum- that is the integral- of all the fields arising from all the dq’s along the rod: Example (Solution)

In order to evaluate the integral we must relate dq to y. The rod carries a uniform charge Q distributed over the length . The line charge density is therefore:

This is the charge per unit length. Thus a length dy carries charge dq given by: Example (Solution)

Does this answer make sense? Example (Solution) Consider

T h e r o d a p p e a r s a s a s m a l l concentration of total charge Q at a distance a away from the point P

END Example (Question) A long straight power line coincides with the x axis and carries a linear charge density ρl C/m. What is the electric field at point P on the y axis. Use the approximation that the line is infinitely long. Here both the direction and magnitude of the electric field element dE arise from charge elements on the line vary with the position x of the charge element.

The figure shows that charge elements on opposite sides of the y axis give rise to electric fields whose x components cancel. Thus the net field points in the y direction, that is away from the positively charged rod. Each element of charge dq contributes an amount dEy to the net electric field at P. Each element of charge dq contributes an amount dEy to the net electric field at P.

We integrate from x = -∞ to x = +∞ x limits Since the line is infinite in both directions and has cylindrical symmetry, the expression for the electric field holds for any point a distance y from the line. The result thus shows that the electric field from a positively charged infinite line points radially away from the line.

The magnitude of the E field drops as 1/y

END Example (Question) A thin ring of radius a is centered on the origin and carries a total charge Q distributed uniformly around the ring. Find the electric field at a point P located a distance x along the axis of the ring, and show that the result makes sense at x >> a. A point on the ring axis is equidistant from all points on t h e r i n g , s o t h e f i e l d magnitudes dE are the same but their directions vary.

The figure shows that any components perpendicular to the x axis cancel for any pair of charge elements on opposite sides of the ring, leaving a net field in the x direction. Each charge element dq contributes an amount dEx to the total field.

with

Gives In this example k, x, and a are constants so we have:

Does this result make sense? At a large distance from the ring x >> a.

The ring appears as a small concentration of total charge Q at a distance x from the point P.

END Example (Question) A s p h e r i c a l s h e l l o f r a d i u s R i s u n i f o r m l y charged with surface charge density ρs. Find the electric field at an exterior point.

Uniform surface charge on shell. dA (x1,y1,z1) P (x2,y2,z2)

here are constants and variables that depend on location of dA. dA P (x2,y2,z2) (x1,y1,z1)

variable that depends on location of dA. dA P (x2,y2,z2) (x1,y1,z1)

variable that depends on location of dA. R

variable that depends on location of dA. Field at P produced by charge element dq:

R 2   ρsR sin(θ)dθdφ E = dE = k 3 [(x2 − x1),(y2 − y1),(z2 − z1)] ∫ ∫ 2 2 2 2 Surface Surface (x2 − x1) + (y2 − y1) + (z2 − z1) ( )

This is not an easy task. It can be done but requires a lot of € work. It can best be solved by converting from (x,y,z) coordinates to (r,θ,φ) coordinates.

END Example (Question) A cylindrical volume of radius R is uniformly charged with volume charge density ρV. Find the electric field at an exterior point.

Uniform volume charge in cylinder. dV (x1,y1,z1)

P (x2,y2,z2)

here are constants and variables that depend on location of dV. (x1,y1,z1) dV P (x2,y2,z2)

variable that depends on location of dV. (x ,y ,z ) dV 1 1 1 P (x2,y2,z2)

variable that depends on location of dV. z

variable that depends on location of dV. P

 ρV rdrdφdz dE = k 3 [(x2 − x1),(y2 − y1),(z2 − z1)] 2 2 2 2 (x2 − x1) + (y2 − y1) + (z2 − z1) ( )

€   ρV rdrdφdz E = dE = k 3 [(x2 − x1),(y2 − y1),(z2 − z1)] ∫ ∫ 2 2 2 2 Volume Volume (x2 − x1) + (y2 − y1) + (z2 − z1) ( )

This is not an easy task. It can be done but requires a lot of work. It can best be solved by converting from (x,y,z) € coordinates to (r,φ,z) coordinates.

END