# Give some examples of electric dipole with explaination

1. Two charges of magnitudes +q and –q are placed at a distance 2a apart. Calculate the electric field at a point distant r from the midpoint of the line joining the two charges.

Electric field at P, E = E1 + E2

E = E1 – E2

$=frac{1}{4pi {{in }_{0}}}left[ frac{q}{{{left( r-a right)}^{2}}}-frac{q}{{{left( r+a right)}^{2}}} right]$ $=left[ frac{1}{4pi {{in }_{0}}}left( frac{2qtimes 2atimes r}{{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}} right) right.$ along Px

But q × 2a = p electric dipole moment

$E=frac{1}{4pi {{in }_{0}}}left[ frac{2pr}{{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}} right]$ along Px

When r > a

Electric field at P,

$E=frac{1}{4pi {{in }_{0}}}left( frac{2p}{{{r}^{3}}} right)$ along Px

1. Two charges of magnitude +q and –q are placed at (+a, 0, 0) and (-a, 0, 0) respectively. Calculate the electric field at a point P whose coordinates are (0, r, 0).

The electric field at P = E1 + E2

${{E}_{1}}={{E}_{2}}=frac{1}{4pi {{varepsilon }_{0}}}left[ frac{q}{left( {{r}^{2}}+{{a}^{2}} right)} right]$

The vector sum of E1 and E2 points along PC,

$E=2{{E}_{1}}cos theta$ … (i)

$cos theta =-frac{a}{sqrt{{{r}^{2}}+{{a}^{2}}}}$

Substituting the values of E and cos θ, in equation (i).

$E=frac{2}{4pi {{in }_{0}}}left[ frac{q}{left( {{r}^{2}}+{{a}^{2}} right)} right]left( -frac{a}{{{left( {{r}^{2}}+{{a}^{2}} right)}^{1/2}}} right)$ $=frac{1}{4pi {{varepsilon }_{0}}}left[ frac{qtimes 2a}{{{left( {{r}^{2}}+{{a}^{2}} right)}^{3/2}}} right]$

Or $E=frac{1}{4pi {{varepsilon }_{0}}}left[ frac{p}{{{left( {{r}^{2}}+{{a}^{2}} right)}^{3/2}}} right]$ along PC

When r > a

$E=frac{1}{4pi {{varepsilon }_{0}}}left( frac{p}{{{r}^{2}}} right)$ along PC.

1. Two charges +20 μC and -20 μC are placed 1 cm apart. Calculate the electric field at a point on the axial line at a distance of 20 cm from the centre of the dipole.

Here dipole moment, p = q × 2a

Here q = 20 μC

2a = 10-2 m

$p=20times {{10}^{-6}}times {{10}^{-2}}=2times {{10}^{-7}}C-m$

r = 0.2 m

$E=frac{1}{4pi {{varepsilon }_{0}}}left( frac{2p}{{{r}^{3}}} right)$ $E=9times {{10}^{9}}left[ frac{2times 2times {{10}^{-7}}}{{{left( 0.2 right)}^{3}}} right]$

E = 4.5 × 105 N/C

The direction of the field is along the axial line.

1. Two charges +20 μC and -20 μC are plaed 1 cm apart. Calculate the electric field at a point on the equatorial line at a distance of 20 cm from the centre of the dipole.

p = q × 2a

here, q = 20 × 10-6 C

2a = 10-2 m

p = 20 × 10-6 × 10-2 = 2 × 10-7 C-m

r = 0.2 m

$E=frac{1}{4pi {{varepsilon }_{0}}}left[ frac{P}{{{r}^{3}}} right]$ $E=9times {{10}^{9}}left[ frac{2times {{10}^{-7}}}{{{left( 0.2 right)}^{3}}} right]$

E = 2.25 × 105 N/C

The direction of the electric field is parallel to the line joining the two charges.

Note:

The dipole moment vector is along a line from –q to +q

Therefore in this case the electric field is parallel and opposite to the direction of dipole moment vector.

1. Two electric charges q and +4q are placed at a distance of 6a apart on a horizontal plane. Find the locus of the point on the line joining the two charges where the electric field is zero.

Let the point be at a distance x from the charge q

Here $frac{q}{4pi {{in }_{0}}{{x}^{2}}}=frac{4q}{4pi {{in }_{0}}{{left( 6a-x right)}^{2}}}$ $therefore frac{1}{{{x}^{2}}}=frac{4}{{{left( 6a-x right)}^{2}}}$

Taking square root

$frac{1}{x}=frac{2}{left( 6a-x right)}$

Or x = 2a

The point is at a distance of 2a from charge q and at a distance of 4a from charge 4q.