Charge and electric flux

Charge and electric flux

Charge & Electric flux

• In last chapter we asked the question "given a charge distribution what is the electric field produced by the distribution at a point P

• The answer can be found by representing the distribution as an assembly of point charges each of which produce an electric field E which is given by the equation


• There is an alternative relationship between charge distribution and electric field

• If electric field pattern is known in the in a given region can we determine about the charge distribution in that region?


• Imagine a box as shown in fig.

• It has unknown amount of charge

• We refer to box as a closed surface

• How can we determine how much electric charge lies within the box?


• Knowing that a charge distribution produces an electric field and that an electric field exerts a force on a test charge.

• We move a test charge q

0 around the vicinity of the box

• By measuring the force F experienced by a test charge at different point


• We make a three dimensional map of the electric field E=F/q

0 outside the box


• To determine the contents of a box we actually measure E on the surface of the box

• Fig. a shows a single +ve point charge inside the box

• Fig b. shows there are two such charges


• The field pattern on the surfaces of the boxes are different in detail but in both cases the electric field points out of the box


• Fig c show and fig. d shows cases with one and two –ve point charges respectively inside the box

• Again the details of E on the surface of the box are different , but in both cases the field points into the box


• The fig. in which electric field pointed outward of the surface the electric flux is outward

• In fig. c in which the electric field points into surface. The electric flux is inward

• Relationship : +ve charge inside the box goes with an outward flux through the box’s surface and –ve charge inside goes with an inward flux


• What happened if there is zero charge inside the box?

• As shown in the fig. a there is no E so there is no electric flux inside or outward

• In fig b. box containing one +ve charge and one – ve charge inside the box with same magnitude


• The net charge inside the box is zero. So there is no electric flux inside the box


• The box is again empty however there is a charge present outside the box the box has been placed with one end parallel to a uniform charged infinite sheet which is produced an electric field perpendicular to the sheet


• One end of the box E points into the box on the opposite end E points out the box

• The inward electric flux exactly compensates for the outward flux on the other part there is no net flux in the surface and there is no net charge


• There is a relationship between magnitude of the net charge inside the closed surface and strength of the net flow of E over the surface

• This suggest that electric flux through the surface of the box is directly proportional to the magnitude of the net charge enclosed by the box


• We have seen that there is relationship between the net amount of charge inside a closed surface and the electric flux through the surface

• Flux through a closed surface depends on the sign of the enclosed charge

• Charge outside the surface do not give a net electric flux through the surface.


• The net electric flux is directly proportional to the amount of the charge enclosed within the surface but is otherwise independent of the size of the closed surface

Calculating Electric flux

• Electric flux through a surface is a description of whether the electric field E points into or out of the surface

• The net electric flux is directly proportional to the net electric charge inside the surface


• Fig. shows a fluid flowing from left to right. Let’s examine the volume flow rate dV/dt through the wire rectangle with area A when the area A is perpendicular to flow of velocity V and the flow is the same at all points in the fluid the volume flow rate dV/dt is the area A multiplied by the flow speed v

• dV/dt = vA


• When the rectangle is tilted at an angle

• So that face is not perpendicular to V

• This area which is outlined in red and labeled A in fig. b is the projection of area A onto a surface perpendicular to V


• Thus the volume flow rate through A is

• dV/dt = vAcosφ

• if φ = 90, dv/dt =0 the wire is edge on the flow and no fluid pass through the rectangle


• Also v cosφ is the component of the vector v perpendicular to the area. Calling this component v we can write the volume rate as


• Using the analogy between electric field and fluid flow, we can define the electric flux in the same way as we have just defined the volume flow rate of a fluid. We simply replace the fluid velocity v by the electric field E

• The symbol that we use for electric flux is Φ

E


• Consider a flat area A perpendicular to a uniform electric field E as in the fig.

• We define the electric flux through this area to be the product of the field magnitude E and the area A

• Φ

E

= EA


• Roughly speaking we can picture the Φ

E in terms of the field lines passing through A. increasing the area means that more lines of E pass through the area, increasing the flux

• Stronger field mean more closely spaced lines of E and therefore more lines per unit area so again the flux increases


• If the area A is flat but not perpendicular to the field E then fewer lines pass through it

• In that case electric flux is given by

• Φ

E

= EAcosΦ


• With a closed surface we will always choose the direction of n to be outward we will speak the flux out of a closed surface thus we called outward flux

• Corresponding to a +ve value of Φ

E and what we called inward flux corresponding to –ve value of Φ

E


• What happened if the electric field is not uniform but varies from point to point over the area A

• Or what happened if A is part of a curved surface?

• Then we divided the A into many small elements dA each of which has a unit vector n perpendicular to it and a vector area

• dA=ndA


• In this case we calculate the electric flux Φ

E

Gauss’s Law

• Gauss’s law says that the total electric flux in any closed surface is proportional to the total electric charge inside the surface

• We’ll start with the field of a single +ve point charge q. the field lines radiate out equally in all direction. We place this charge at the center of an imaginary spherical surface with radius R. The magnitude E of the electric field at every point on the surface is given by

Gauss’s Law

• At each point on the surface E is perpendicular to the surface and its magnitude is the same at every point . The total electric flux is just the product of the field magnitude E and the total area A = 4πR 2 of the sphere

• Φ

E

= EA = 1/4πЄ

0

.q/ R 2 (4πR 2 ) = q/Є

0

• The flux is independent of the radius R of the sphere. It depends only on the charge q enclosed by the sphere

Gauss’s Law

• We can divide the entire irregular surface into element dA compute the electric flux EAcosΦ for each and sum the result by integrating

Gauss’s Law

• This equation holds for any surface of any shape and size, provided only that it is a closed surface enclosing the charge q. the circle on the integral sign reminds us that integral is always closed surface

Gauss’s Law

• Now comes the final step in obtaining the general form of gauss’s law.

• Suppose the surface encloses not just one point charge q but serval charges q1,q2,q3…qn

• The total electric field E at any point is the vector sum of the E field of the individual charges

Gauss’s Law

• Let Qenc be the total charge enclosed by the closed surface Qenc = q1+q2+a7q3…..

• Also let E be the total field at the position of the surface area element dA and let E be its component perpendicular to the plane of that element. Then we can write general statement of gauss’s law

Charge and electric flux

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