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Electric Charges and Fields

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
Show that electric field due to line charge at any plane is same in magnitude and directed radially upward.

A thin long straight line L of charge having uniform linear charge density λ is shown by figure. By symmetry, it follows that electric field due to line charge at distance r in any plane is same in magnitude and directed radially upward.

A thin long straight line L of charge having uniform linear charge de

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A thin conducting spherical shell of radius R has charge +q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell. Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.

Electric field intensity at any point outside a uniformly charged spherical shell:
Assume a thin spherical shell of radius R with centre O. Let charge +q is uniformly distributed over the surface of the shell.

Electric field intensity at any point outside a uniformly charged sph
Let P be any point on the Gaussian surface sphere S1 with centre O and radius r (r > R). According to Gauss's law

Electric field intensity at any point outside a uniformly charged sph
Graph: As charge on shell reside on outer surface so there is no charge inside shell so electric field by Gauss's law will be zero.

Electric field intensity at any point outside a uniformly charged sph


The variation of the electric field intensity E(r) with distance r from the centre for shell 0 ≤ r < ∞ is shown below.

Electric field intensity at any point outside a uniformly charged sph


  

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What is meant by the statement that the electric field of a point charge has spherical symmetry whereas that of an electric dipole is cylindrically symmetrical?

Consider a charge q at the centre of a sphere of radius r. The electric field at all points on the surface of the sphere is given by

Consider a charge q at the centre of a sphere of radius r. The electr
So, the electric field due to a point charge is spherically symmetric.
In the case of an electric dipole, the electric field at a distance r, from the mid-point of the dipole, on the equatorial line is given by

Consider a charge q at the centre of a sphere of radius r. The electr
Now, imagine a cylinder of radius r drawn with electric dipole as axis. The electric field, due to dipole, at all points on the surface of the cylinder will be the same. So, the electric field due to dipole has cylindrical symmetry.

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State Gauss's law in electrostatics. Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density A cm–1.

Gauss’s law in electrostatics: It states that total electric flux over the closed surface S in vacuum is 1 over straight epsilon subscript 0 times the total charge (q) contained in side S.
therefore                     straight ϕ subscript straight E space equals space contour integral for straight s of straight E with rightwards arrow on top. space ds with rightwards arrow on top space equals space straight q over straight epsilon subscript 0
Electric field due to an infinitely long straight wire.
Let an infinitely long line charge having linear charge density X. Assume a cylindrical Gaussian surface of radius r and length 1 coaxial with the line charge to determine its electric field at distance r.
By symmetry, the electric field E has same magnitude at each point of the curved surface S1 and is directed radially outward. So angle at surfaces between dS with rightwards arrow on top space and space straight E with rightwards arrow on top is zero, and angle of stack dS subscript 2 with rightwards arrow on top comma space space stack dS subscript 3 with rightwards arrow on top with straight E with rightwards arrow on top at straight S subscript 2 space and space straight S subscript 3 are 90 degree.

Gauss’s law in electrostatics: It states that total electric flux o
Total flux through the cylindrical surface,
contour integral straight E with rightwards arrow on top. ds with rightwards arrow on top space equals space contour integral for straight S subscript 1 of straight E with rightwards arrow on top. space stack dS subscript 1 with rightwards arrow on top space plus space contour integral for straight S subscript 2 of stack straight E. with rightwards arrow on top space stack dS subscript 2 with rightwards arrow on top plus space contour integral for straight S subscript 3 of straight E with rightwards arrow on top. space stack dS subscript 3 with rightwards arrow on top

space equals space contour integral for straight S subscript 1 of EdS subscript 1. space cos space 0 degree space plus space contour integral for straight S subscript 2 of EdS subscript 2. space cos space 90 degree plus space contour integral for straight S subscript 3 of EdS subscript 3. space cos space 90 degree
equals space straight E space contour integral space dS subscript 1 space equals space straight E space cross times space 2 πrl
Since λ is the charge per unit length and l is the length of the wire, Thus, the charge enclosed
q = λl
According to Gaussian law,

Gauss’s law in electrostatics: It states that total electric flux o
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The electrostatic force of repulsion between two positively charged ions carrying equal charge is 3.7 x 10–9 N, when they are separated by a distance of 5A . How many electrons are missing from each ion?

Here given,
           Electrostatic force of repulsion -F = 3.7 × 10-9N

 Let us say charge is q1 = q2 = q distance between two charges-r = 5 Å = 5 × 10-10 m,To find the number of electrons missing - n = ?

Using Coulomb's law,
                   
                  F = 14π 0q1 q2r2
         
      3.7 × 10-9 = 9 × 109 qq(5 × 10-10)2          q2 = 3.7 × 10-9 × 25 × 10-209 × 109                    = 10.28 × 10-38            q = 3.2 × 10-19 coulomb

As           q = ne
            n=qe=3.2 × 10-191.6 × 10-19 = 2

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A charge q is placed at the centre of the line joining two equal charges Q. Show that the system of three charges will be in equilibrium if q = – Q / 4.

Let two equal charges Q each, be held at A and B, where AB = 2x. C is the centre of AB, where charge q is held, figure below.

Let two equal charges Q each, be held at A and B, where AB = 2x. C is
Net force on q is zero. So q is already in equilibrium.
For the three charges to be in equilibrium, net force on each charge must be zero. Now, total force on Q at B is
              fraction numerator 1 over denominator 4 straight pi space element of subscript 0 end fraction fraction numerator straight Q space straight q over denominator straight x squared end fraction plus fraction numerator 1 over denominator 4 straight pi element of subscript 0 end fraction fraction numerator QQ over denominator left parenthesis 2 straight x right parenthesis squared end fraction space equals space 0
or                      fraction numerator 1 over denominator 4 straight pi space element of subscript 0 end fraction Qq over straight x squared space equals space minus fraction numerator 1 over denominator 4 straight pi element of subscript 0 end fraction fraction numerator straight Q space straight Q over denominator left parenthesis 2 straight x right parenthesis squared end fraction space equals space 0
or                straight q equals negative straight Q over 4
which was to be proved.
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