Physics, asked by moni3735, 1 year ago

Calculate the electric field due to an electric dipole of length 10cm having charge of one micro coulomb at an equatorial point 12cm from the center of the diole

Answers

Answered by abhi178
35

Therefore the Electric field due to an electric dipole is 4.1 × 10^5 N/C.

Given : An electric dipole of length 10cm having charge of one micro coulomb at an equatorial point 12cm from the centre of the dipole.

To find : The electric field due to the electric dipole.

solution : The electric field due to an electric dipole is given by, E = KP/(r² + a²)³/²

where P = q2a = 10^-6 C × 10 × 10^-2 m = 10^-7 Cm

r = 12 cm = 0.12 m and a = 5cm = 0.05 m

The electric field, E = (9 × 10^9 × 10^-7)/[(0.12)² + (0.05)²]³/²

= (900)/[0.0144 + 0.0025]³/²

= 900/(0.0169)³/²

= 900/(0.13)³

= 900/0.002197

= 409,649.522

≈ 4.1 × 10^5 N/C

Therefore the Electric field due to an electric dipole is 4.1 × 10^5 N/C.

Answered by Anonymous
4

Answer:

Therefore the Electric field due to an electric dipole is 4.1 × 10^5 N/C.

Given :

An electric dipole of length 10cm having charge of one micro coulomb at an equatorial point 12cm from the centre of the dipole.

To find :

The electric field due to the electric dipole.

solution :

The electric field due to an electric dipole is given by, E = KP/(r² + a²)³/²

where P = q2a = 10^-6 C × 10 × 10^-2 m = 10^-7 Cm

r = 12 cm = 0.12 m and a = 5cm = 0.05 m

The electric field, E = (9 × 10^9 × 10^-7)/[(0.12)² + (0.05)²]³/²

= (900)/[0.0144 + 0.0025]³/²

= 900/(0.0169)³/²

= 900/(0.13)³

= 900/0.002197

= 409,649.522

≈ 4.1 × 10^5 N/C

Therefore the Electric field due to an electric dipole is 4.1 × 10^5 N/C.

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