Electric Charges and Fields

Physics Part I

Physics

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

Consider a thin long straight line L of charge having uniform linear charge density λ (as shown in fig.)

The direction of Electric field from point O of the line charge is directed radially outwards.

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.

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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 5$\stackrel{\circ}{\mathrm{A}}$ . How many electrons are missing from each ion?

Here given,

$\mathrm{Electrostatic}\mathrm{force}\mathrm{of}\mathrm{repulsion}-\mathrm{F}=3.7\times {10}^{-9}\mathrm{N}$

$\mathrm{Let}\mathrm{us}\mathrm{say}\mathrm{charge}\mathrm{is}{\mathrm{q}}_{1}={\mathrm{q}}_{2}=\mathrm{q}\phantom{\rule{0ex}{0ex}}\mathrm{distance}\mathrm{between}\mathrm{two}\mathrm{charges}-r=5\mathrm{\AA}=5\times {10}^{-10}\mathrm{m},\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{electrons}\mathrm{missing}-\mathrm{n}=?$

Using Coulomb's law,

$\mathrm{F}=\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}$

$\Rightarrow 3.7\times {10}^{-9}=9\times {10}^{9}\frac{\mathrm{qq}}{(5\times {10}^{-10}{)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{q}}^{2}=\frac{3.7\times {10}^{-9}\times 25\times {10}^{-20}}{9\times {10}^{9}}\phantom{\rule{0ex}{0ex}}=10.28\times {10}^{-38}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{q}=3.2\times {10}^{-19}\mathrm{coulomb}$

As $\mathrm{q}=\mathrm{ne}$

$\therefore $ $\mathrm{n}=\frac{\mathrm{q}}{\mathrm{e}}=\frac{3.2\times {10}^{-19}}{1.6\times {10}^{-19}}=2$

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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.

Let P be any point on the Gaussian surface sphere S

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.

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

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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

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

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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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.

Net force acting 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 due to q and Q at A is

$\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{\mathrm{Q}\mathrm{q}}{{\mathrm{x}}^{2}}+\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{\mathrm{QQ}}{(2\mathrm{x}{)}^{2}}=0$

$\Rightarrow $ $\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{\mathrm{Qq}}{{\mathrm{x}}^{2}}=-\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{\mathrm{Q}\mathrm{Q}}{(2\mathrm{x}{)}^{2}}=0$

$\Rightarrow $ $\mathrm{q}=-\frac{\mathrm{Q}}{4}$

Hence proved.

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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 is $\frac{1}{{\mathrm{\epsilon}}_{0}}$ times the total charge (q) contained in side S.

$\therefore $ $\overline{){\mathrm{\varphi}}_{\mathrm{E}}=\underset{\mathrm{s}}{\oint}\overrightarrow{\mathrm{E}}.\overrightarrow{ds}=\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{0}}}$

Electric field due to an infinitely long straight wire.

Let us consider an infinitely long line charge having linear charge density $\mathrm{\lambda}$. Assume a cylindrical Gaussian surface of radius r and length 1 coaxial with the line charge.

By symmetry, the electric field E has same magnitude at each point of the curved surface S

Total flux through the cylindrical surface,

$\oint \overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{ds}}=\underset{{\mathrm{S}}_{1}}{\oint}\overrightarrow{\mathrm{E}}.\overrightarrow{{\mathrm{dS}}_{1}}+\underset{{\mathrm{S}}_{2}}{\oint}\overrightarrow{\mathrm{E}.}\overrightarrow{{\mathrm{dS}}_{2}}+\underset{{\mathrm{S}}_{3}}{\oint}\overrightarrow{\mathrm{E}}.\overrightarrow{{\mathrm{dS}}_{3}}$

$=\underset{{\mathrm{S}}_{1}}{\oint}{\mathrm{EdS}}_{1}.\mathrm{cos}0\xb0+\underset{{\mathrm{S}}_{2}}{\oint}{\mathrm{EdS}}_{2}.\mathrm{cos}90\xb0+\underset{{\mathrm{S}}_{3}}{\oint}{\mathrm{EdS}}_{3}.\mathrm{cos}90\xb0\phantom{\rule{0ex}{0ex}}=\mathrm{E}\oint {\mathrm{dS}}_{1}=\mathrm{E}\times 2\mathrm{\pi rl}$

Since λ is the charge per unit length and l is the length of the wire.

Thus, the charge enclosed

$=\underset{{\mathrm{S}}_{1}}{\oint}{\mathrm{EdS}}_{1}.\mathrm{cos}0\xb0+\underset{{\mathrm{S}}_{2}}{\oint}{\mathrm{EdS}}_{2}.\mathrm{cos}90\xb0+\underset{{\mathrm{S}}_{3}}{\oint}{\mathrm{EdS}}_{3}.\mathrm{cos}90\xb0\phantom{\rule{0ex}{0ex}}=\mathrm{E}\oint {\mathrm{dS}}_{1}=\mathrm{E}\times 2\mathrm{\pi rl}$

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,

According to Gaussian law,

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