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Question

(a) Deduce the expression for the potential energy of a system of two charges q1 and q2 located r 1 and r 2 , respectively, in an external electric field.

(b) Three point charges, + Q + 2Q and – 3Q are placed at the vertices of an equilateral triangle ABC of side l. If these charges are displaced to the mid-point A1, B1 and C1 ,respectively, find the amount of the work done in shifting the charges to the new locations.

OR

Define electric flux. Write its S.I unit.

State and explain Gauss’s law. Find out the outward flux to a point charge +q placed at the centre of a cube of side ‘a’. Why is it found to be independent of the size and shape of the surface enclosing it? Explain.

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Solution

The work done in bringing the chare q1 from infinity to r2 can be calculated. Here, the work is done not only against the external field E but also against the field due to q1 Hence, work done on q2 against the external field is W2=q2V(r2)

Work done on q against the field due to q1 ,W12=q1q24πϵ0r12

where r12 is the distance between q1 and q2

By the principle of superposition for fields, work done on q2 against two fields will add with work done in bringing q2 to r2 , which is given as

W+W=qV(r2)+q1q24πϵ0r12

Thus, the potential energy of the system U = total work done in assembling the configuration U

W1+W2+W+12

U=qV(r1)+qV(r2)+q1q24πϵ0r12

(b)

q=+Q

q = +2Q

q = -3Q

r = l (for each side)

Intial potential energy of system

U1=14πϵ01[(q1×q2)+(q2×q3)+(q3×q1)]U1=14πϵ01[(+Q×+2Q)+(+2Q×3Q)+(3Q×+Q)+]U1=7Q24πϵ01

These charges displaced to mid points then final potential energy of system

U1=14πϵ01/2[(q1×q2)+(q2×q2)+(q3×q2)]U2=24πϵ01[(+Q×+2Q)+(2Q×3Q)+(3Q×+Q)]U2=7Q24πϵ01

Work done, W=U2U1W=7Q22πϵ017Q24πϵ01W=7Q2πϵ01[12(14)]=7Q2πϵ01[12+14]W=74(Q2πϵ01)

OR

Electric flux is the total number of lines of force passing through the unit area of a surface held perpendicularly.

Electric fluxΔϕ through an area element ΔS is given by

Δϕ=E.ΔS=EΔScosθ

θis the angle between E and ΔS

The SI unit of electric flux is NC1m2

Gauss’s law states that the flux of the electric field through any closed surface S is 1/ε0 times the total charge enclosed by S.

Let the total flux through a sphere of radius r enclose a point charge q at its centre. Divide the sphere into a small area element as shown in the figure.

The flux through an area element ΔS is

Δϕ=EΔS=q4πϵ0r2^r.ΔS

Here, we have used Coulomb’s law for the electric field due to a single charge q.

The unit vector ^r is along the radius vector from the centre to the area element. Because the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and ^r have the same direction. Therefore,

Δϕ=q4πϵ0r2.ΔS

Because the magnitude of the unit vector is 1, the total flux through the sphere is obtained by adding the flux through all the different area elements.

ϕ=allΔS=q4πϵ0r2.ΔS

Because each area element of the sphere is at the same distance r from the charge,

ϕ=q4πϵ0r2allΔS=q4πϵ0r2.S

Now, S the total area of the sphere equals 4πr2 .Thus

ϕ=q4πϵ0r2×4πr2=qϵ0

Hence, the above equation is a simple illustration of a general result of electrostatics called Gauss’s law.

Let a cube of side a enclose charge +q at its centre.

Because the electric flux through the square surface isϕ=q6ϵ0 the square surfaces of cube are six. Hence, according to Gauss’s theorem in electrostatics, the total outward flux due to a charge +q of a cube is ϕ=6×(q6ϵ0)=qϵ0The result shows that the electric flux passing through a closed surface is proportional to the charge enclosed. In addition, the result reinforces that the flux is independent of the shape and size of the closed surface.


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