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# Electric Field Assignment Help

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Electromagnetism - Electric Field

**Electric Field**

The clods integral of electric field intensity is equal to q/ε0 where q is charge enclosed in the closed surface. In other words total electric flux through a closed surface enclosing a charge q is given by **§E. ****dS = q/ε**_{0}

If W is at right angle to the surface area A at all points and has the same magnitude at all points of the surface then E_{1}** = E and ∫E**_{1} d_{A} = E_{A}.

If E is parallel to the surface on all points then **E**_{1} = 0 hence integral is also zero.

If **E = 0** hence integral is also zero.

If **E = 0** at all points on a surface then **∅ = 0**.

The surface need not be a real physical surface; it can be a hypothetical one.

Electric field in**?E. dS** is complete electric field. It may be partly far to charge within the surface and partly due to charge courtside the surface. However, if there is no charge enclosed in the Gaussian surface **E**_{1} will be zero and hence **?E dS = 0**.

While evaluating**?E, dS**, the field should lie on the surface and there should be enough symmetry to evaluate the integral.

Electric field due to a long threads (Line charge) having linear charge density** λ** is

**E = λ/2πε**_{0}y = 18 x 10^{9} λ/y

Electric field due to a uniformly charged sphere of radius R having charge Q

**E **_{inside} = Qx/4πε_{0} R^{3} x < R

E _{surface} = Q/4πε_{0}R x = R

E _{outside }= Q/4πε_{0} x^{2} x > R

Potential due to a uniformly charged sphere

**V**_{ inside} = Q/4πε_{0}R + ∫xR - Qx/4πε_{0}R^{3} dx x < R

V _{surface} = Q/4πε_{0}R x = R

V _{outside} = Q/4πε_{0}R x > R

Electric filed doer to a thin plane sheet (long) of charge density **σ**

**E = σ/2ε**_{0}

Electric field due to a charged surface having surface charge density **σ**

**E = σ/ε**_{0}

Electric field due to a conducting plate:- **E = σ/2ε**_{0}

Electric field due to a non-conducting plate:- ** E = σ/ε0**

Electric field between two oppositely charged sheets at any point is **E**_{in} = σ/ε_{0} = (E_{1} + E2) assuming equal surface charge density, for example, in a capacitor. Electric field intensity is zero at any point outside the plates as **E**_{net} = E_{1} - E_{2} = 0.

ExpertsMind.com - Electric Field Assignment Help, Electric Field Homework Help, Electric Field Assignment Tutors, Electric Field Solutions, Electric Field Answers, Electromagnetism Assignment Tutors

**Electric Field**

**§E.**

**dS = q/ε**

_{0}

If W is at right angle to the surface area A at all points and has the same magnitude at all points of the surface then E

_{1}

**= E and ∫E**

_{1}d_{A}= E_{A}.If E is parallel to the surface on all points then

**E**hence integral is also zero.

_{1}= 0If

**E = 0**hence integral is also zero.

If

**E = 0**at all points on a surface then

**∅ = 0**.

The surface need not be a real physical surface; it can be a hypothetical one.

Electric field in

**?E. dS**is complete electric field. It may be partly far to charge within the surface and partly due to charge courtside the surface. However, if there is no charge enclosed in the Gaussian surface

**E**

_{1}will be zero and hence

**?E dS = 0**.

While evaluating

**?E, dS**, the field should lie on the surface and there should be enough symmetry to evaluate the integral.

Electric field due to a long threads (Line charge) having linear charge density

**λ**is

**E = λ/2πε**

_{0}y = 18 x 10^{9}λ/yElectric field due to a uniformly charged sphere of radius R having charge Q

**E**

E

E

_{inside}= Qx/4πε_{0}R^{3}x < RE

_{surface}= Q/4πε_{0}R x = RE

_{outside }= Q/4πε_{0}x^{2}x > RPotential due to a uniformly charged sphere

**V**

V

V

_{ inside}= Q/4πε_{0}R + ∫xR - Qx/4πε_{0}R^{3}dx x < RV

_{surface}= Q/4πε_{0}R x = RV

_{outside}= Q/4πε_{0}R x > RElectric filed doer to a thin plane sheet (long) of charge density

**σ**

**E = σ/2ε**

_{0}

Electric field due to a charged surface having surface charge density

**σ**

**E = σ/ε**

_{0}

Electric field due to a conducting plate:-

**E = σ/2ε**

_{0}

Electric field due to a non-conducting plate:-

**E = σ/ε0**

Electric field between two oppositely charged sheets at any point is

**E**assuming equal surface charge density, for example, in a capacitor. Electric field intensity is zero at any point outside the plates as

_{in}= σ/ε_{0}= (E_{1}+ E2)**E**.

_{net}= E_{1}- E_{2}= 0ExpertsMind.com - Electric Field Assignment Help, Electric Field Homework Help, Electric Field Assignment Tutors, Electric Field Solutions, Electric Field Answers, Electromagnetism Assignment Tutors