Two charges ±10 µC are placed 5.0 mm apart.Determine the electric fi

1.	Two charges ±10 µC are placed 5.0 mm apart.Determine the electric field at (a) a point P on the axis of the dipole15 cm away from its centre O on the side of the positive charge, asa point Q, 15 cm away from O on a linepassing through O and normal to the axis of the dipole,
Answer» answer : (i) 2.6 × 10^5 N/C (ii) 1.3 × 10^5 N/C It has given that two charges +10μC and -10μC are seperated 5MM apart. so, dipole, P = qd = 10μC × 5mm = 10 × 10^-6 C × 5 × 10^-3 m = 5 × 10^-8 Cm (i) electric field at a point on axial line, E = 2KP/r³ = (2 × 9 × 10^9 × 5 × 10^-8)/(15 × 10^-2)³ [ here, r = 5cm ] = (90 × 10)/(3375 × 10^-6) = 0.266 × 10^6 = 2.66 × 10^5 N/C (ii) electric field at a point on equatorial line, E = KP/r³ = (9 × 10^9 × 5 × 10^-8)/(5 × 10^-2)³ = (45 × 10)/(3375 × 10^-6) = 0.133 × 10^6 = 1.33 × 10^5 N/C

Two charges ±10 µC are placed 5.0 mm apart.Determine the electric field at (a) a point P on the axis of the dipole15 cm away from its centre O on the side of the positive charge, asa point Q, 15 cm away from O on a linepassing through O and normal to the axis of the dipole,

Answer»

answer : (i) 2.6 × 10^5 N/C (ii) 1.3 × 10^5 N/C

It has given that two charges +10μC and -10μC are seperated 5MM apart.

so, dipole, P = qd

= 10μC × 5mm

= 10 × 10^-6 C × 5 × 10^-3 m

= 5 × 10^-8 Cm

(i) electric field at a point on axial line, E = 2KP/r³

= (2 × 9 × 10^9 × 5 × 10^-8)/(15 × 10^-2)³ [ here, r = 5cm ]

= (90 × 10)/(3375 × 10^-6)

= 0.266 × 10^6 = 2.66 × 10^5 N/C

(ii) electric field at a point on equatorial line, E = KP/r³

= (9 × 10^9 × 5 × 10^-8)/(5 × 10^-2)³

= (45 × 10)/(3375 × 10^-6)

= 0.133 × 10^6 = 1.33 × 10^5 N/C

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Two charges ±10 µC are placed 5.0 mm apart.Determine the electric field at (a) a point P on the axis of the dipole15 cm away from its centre O on the side of the positive charge, asa point Q, 15 cm away from O on a linepassing through O and normal to the axis of the dipole,

answer : (i) 2.6 × 10^5 N/C (ii) 1.3 × 10^5 N/C

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