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Answer:
at d the charge will be 3q and at 3d it will be 9q
Explanation:
for V=Vp-V2d
V=KQ/d=K*6q/2d=3kq/d for potential to 2d at 6q be zero the Vp will equal 3kq/d; hence at d, Q=3q and at 3d, Q=9q
Light<span> refracts whenever it </span>travels<span> at an angle into a substance with a </span>different<span>refractive index (optical density). This change of direction is </span>caused<span> by a change in speed. For example, when </span>light travels<span> from air into water, it slows down, causing it to continue to travel at a </span>different<span> angle or direction.</span>
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:
The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:
The wavelength of the 2nd harmonic is:
The wavelength of the 4th harmonic is:
It is not possible to find any integer n such that 
, therefore the correct options are A, B and D.
