CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A system has two charges $${q}_{A}=2.5\times {10}^{-7}$$ and $${q}_{B}=2.5\times {10}^{-7}$$, located at point $$A$$: $$(0, 0, -15\ cm)$$, $$B$$: $$(0, 0, 15\ cm)$$respectively. What are the total charge and electric dipole moment of the system?


Solution

Charge on A, $$q_A=2.5 \times 10 ^{-7} C$$
Charge on B, $$q_B=2.5 \times 10 ^{-7} C$$
Therefore, total charge q will be
$$q=q_A+q_B$$
   $$=2.5\times10^{-7}+2.5\times10^{-7}$$
   $$=5\times10^{-7}C$$
Distance between two charges, $$a=15-(-15)=30cm=0.3m$$
Electron dipole moment,
$$p=q.a$$
   $$=2.5\times10^{-7}\times0.3$$
   $$=7.5\times 10^{-7}Cm$$  (along z-axis)
Because the direction of dipole is always negative to positive.

1007522_1024570_ans_17354e7483f5417b98bc329342f2d2a8.png

Physics
NCERT
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image