Question

Solve log_2_(x+1)<1.​

Answer 1

Step-by-step explanation:

Therefore, the ratio of electric field intensity at a point on the equatorial line to the field at a point on axial line

\frac{E_{\text{axial}}}{E_{\text{equatorial}}}=\frac{\frac{1}{4\times\pi\times\varepsilon_{\circ}}\times\frac{2P}{r^{3}}}{\frac{1}{4\times\pi\times\varepsilon_{\circ}}\times\frac{P}{r^{3}}}=2

4.5