Question

two charges each tourist 2 × 10 to the power minus 7 coulomb but opposite sign for my system these charges are located at point A 0 0 - 10 and B 0 0 10 respectively calculate the electric dipole of the system

Answer 1

☆$\red {HELLO\:DEAR}$☆

Both the charges can be located in a coordinate frame of reference as shown in the given figure.

At A, amount of charge, qA = 2.5 × 10−7C

At B, amount of charge, qB = −2.5 × 10−7 C

Total charge of the system,

q = qA + qB

= 2.5 × 10−7 C − 2.5 × 10−7 C

= 0

Distance between two charges at points A and B,

d = 15 + 15 = 30 cm = 0.3 m

Electric dipole moment of the system is given by,

p = qA × d = qB × d

= 2.5 × 10−7 × 0.3

= 7.5 × 10−8 C m along positive z-axis

Therefore, the electric dipole moment of the system is 7.5 × 10−8 C m along positive z−axis.

☆$\green {VISHU\:PANDAT}$☆

☆$\blue {FOLLOW\:ME}$☆

Answer 2

Explanation:

At A, amount of charge,

qa= 2.5×10^−7C (Given)

At B, amount of charge,

qb =−2.5×10^−7C

Total charge of the system,

q=qa +qb

q=2.5×10^7 - 2.5×10^−7

C=0

Distance between two charges at points A and B,

d=10+10=30cm=0.2m

Electric dipole moment of the system is given by,

p=qa ×d

p=2.5×10^−7×0.2

p= 5×10^−8 cm along positive z-axis.

Therefore, the electric dipole moment of the system is 5×10^−8

along the positive z-axis.