Question

Two charges + 3.2 into 10 to the power of minus 19 coulomb and -3.2 into 10 to the power of minus 19 coulomb placed at 2.4 amperes apart from the electric dipole is placed in a uniform electric field of intensity 4 into 10 ^ 5 volt per metre electric dipole moment is

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

7.68 X 10^-²⁹

Explanation:

dipole moment = qx2l

q= 3.2 X 10^-¹⁹

2l= 2.4 X 10^-¹⁰

p= (3.2 X 10^-¹⁹)(2.4 X 10^-¹⁰)=7.68 X 10^-²⁹