Question

The force experienced by a unit charge where placed at a distance of 0.01 m from the middle of an electric dipole on its axial line of 0.025 N amd when it is placed at a distance of 0.2 m, the force is reduced to 0.002 N. Calculate the dipole length.

Answer 1

Answer:

Let length of dipole is 2a then elecrtric field on axial line is given by

E = 2kpr/(r2 – a2)2

Then

E1/E2 = (r1/r2)*((r22 – a2)2/(r12 – a2)2)

\sqrt{E1*r2/E2*r1} = (r2^{2} - a^{2})/(r1^{2} - a^{2})

\sqrt{0.025*0.2/0.002*0.1} = (0.2^{2} - a^{2})/(0.1^{2} - a^{2})

5 = (0.2^{2} - a^{2})/(0.1^{2} - a^{2})

0.05 – 5a2 = 0.04 – a2

0.01 = 4a2

a = 1/20 m = 5 cm

then length = 2a = 10 cm.

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Explanation:

Let length of dipole is 2a then elecrtric field on axial line is given by

E = 2kpr/(r2 – a2)2

Then

E1/E2 = (r1/r2)*((r22 – a2)2/(r12 – a2)2)

\sqrt{E1*r2/E2*r1} = (r2^{2} - a^{2})/(r1^{2} - a^{2})

\sqrt{0.025*0.2/0.002*0.1} = (0.2^{2} - a^{2})/(0.1^{2} - a^{2})

5 = (0.2^{2} - a^{2})/(0.1^{2} - a^{2})

0.05 – 5a2 = 0.04 – a2

0.01 = 4a2

a = 1/20 m = 5 cm

then length = 2a = 10 cm.

Hope it clears If you like answer then please approve it.