Question #126043
What is the strength of an electric field resulting from a dipole with a dipole moment of 25 C⋅m at a radius, r = 30 cm away and 75° from the y-axis?
1
Expert's answer
2020-07-13T11:43:04-0400

Let the direction of dipole is along the positive direction of x-axis.

Since,

E(r)=3(pr^)r^p4πε0r3 .{\displaystyle \mathbf {E} \left(\mathbf {r} \right)={\frac {3\left(\mathbf {p} \cdot {\hat {\mathbf {r} }}\right){\hat {\mathbf {r} }}-\mathbf {p} }{4\pi \varepsilon _{0}r^{3}}}\ .}

And,

p=25i^r=0.3(cos(15)i^+sin(15)j^)\bold p=25\hat{i}\\ \bold r=0.3(\cos(15^\circ)\hat{i}+\sin(15^\circ)\hat{j})

Now.

\bold p\cdot\hat\bold r=25\cos(15^\circ)

Thus,

E(r)=9×1090.33(3(25cos(15))(cos(15)i^+sin(15)j^)25i^    E=10123(44.97i^+18.75j^)    E=16.24×1012N/C\bold E(\bold r)=\frac{9\times 10^9}{0.3^3}(3(25\cos(15^\circ))(\cos(15^\circ)\hat{i}+\sin(15^\circ)\hat{j})-25\hat{i}\\ \implies \bold E=\frac{10^{12}}{3}(44.97\hat{i}+18.75\hat j)\\ \implies E=16.24\times10^{12}N/C


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS