A10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. the electric field between the plates is increasing at the rate 1.5×106 v/ms . what is the magnetic field strength on the axis.

The magnetic field β =  3.1136 x 10 ⁻¹⁴ T

Explanation:

The miss information of r = 6.7 cm

The magnetic field can be find using the formula

β = μ0 ε0 (R² / 2r) dE/dt

Knowing the constant as μ0 and ε0

μ0 = 4π × 10⁻⁷ Tm/A

ε0 = 8.854 × 10^-12 C²/N∙m²

R = 0.050 m

r = 0.067 m

dE/dt = 1.5×10⁵ V/m∙s

β = 4π × 10⁻⁷ Tm/A  * 8.854 × 10⁻¹² C²/N∙m²  * ( 0.050²m / 2* 0.067 m )* 1.5×10⁵ V/m∙s    

β =  3.1136 x 10 ⁻¹⁴ T

d. learned preferences i believe is your answer

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.